Calculus Examples

$f (x) = x^{2} e^{- x}$

Step 1

Consider the limit definition of the derivative.

$f' (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Step 2

Find the components of the definition.

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Step 2.1

Evaluate the function at $x = x + h$ .

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Step 2.1.1

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = {(x + h)}^{2} e^{- (x + h)}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Rewrite ${(x + h)}^{2}$ as $(x + h) (x + h)$ .

$f (x + h) = (x + h) (x + h) e^{- (x + h)}$

Step 2.1.2.2

Expand $(x + h) (x + h)$ using the FOIL Method.

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Step 2.1.2.2.1

Apply the distributive property.

$f (x + h) = (x (x + h) + h (x + h)) e^{- (x + h)}$

Step 2.1.2.2.2

Apply the distributive property.

$f (x + h) = (x \cdot x + x h + h (x + h)) e^{- (x + h)}$

Step 2.1.2.2.3

Apply the distributive property.

$f (x + h) = (x \cdot x + x h + h x + h \cdot h) e^{- (x + h)}$

Step 2.1.2.3

Simplify and combine like terms.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Multiply $x$ by $x$ .

$f (x + h) = (x^{2} + x h + h x + h \cdot h) e^{- (x + h)}$

Step 2.1.2.3.1.2

Multiply $h$ by $h$ .

$f (x + h) = (x^{2} + x h + h x + h^{2}) e^{- (x + h)}$

Step 2.1.2.3.2

Add $x h$ and $h x$ .

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Step 2.1.2.3.2.1

Reorder $x$ and $h$ .

$f (x + h) = (x^{2} + h x + h x + h^{2}) e^{- (x + h)}$

Step 2.1.2.3.2.2

Add $h x$ and $h x$ .

$f (x + h) = (x^{2} + 2 h x + h^{2}) e^{- (x + h)}$

Step 2.1.2.4

Apply the distributive property.

$f (x + h) = (x^{2} + 2 h x + h^{2}) e^{- x - h}$

Step 2.1.2.5

Apply the distributive property.

$f (x + h) = x^{2} e^{- x - h} + 2 h x e^{- x - h} + h^{2} e^{- x - h}$

Step 2.1.2.6

The final answer is $x^{2} e^{- x - h} + 2 h x e^{- x - h} + h^{2} e^{- x - h}$ .

$x^{2} e^{- x - h} + 2 h x e^{- x - h} + h^{2} e^{- x - h}$

Step 2.2

Reorder.

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Step 2.2.1

Reorder $- x$ and $- h$ .

$x^{2} e^{- h - x} + 2 h x e^{- x - h} + h^{2} e^{- x - h}$

Step 2.2.2

Reorder $- x$ and $- h$ .

$x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- x - h}$

Step 2.2.3

Reorder $- x$ and $- h$ .

$x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x}$

Step 2.3

Find the components of the definition.

$f (x + h) = x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x}$

$f (x) = x^{2} e^{- x}$

$f (x + h) = x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x}$

$f (x) = x^{2} e^{- x}$

Step 3

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - (x^{2} e^{- x})}{h}$

Step 4

Remove parentheses.

$f' (x) = lim_{h \to 0} \frac{x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - x^{2} e^{- x}}{h}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 5.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{h \to 0} x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{lim_{h \to 0} x^{2} e^{- h - x} + lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.2

Move the term $x^{2}$ outside of the limit because it is constant with respect to $h$ .

$\frac{x^{2} lim_{h \to 0} e^{- h - x} + lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.3

Move the limit into the exponent.

$\frac{x^{2} e^{lim_{h \to 0} - h - x} + lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.4

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - lim_{h \to 0} x} + lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.5

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.6

Move the term $2 x$ outside of the limit because it is constant with respect to $h$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.7

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot lim_{h \to 0} e^{- h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.8

Move the limit into the exponent.

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{lim_{h \to 0} - h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.9

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - lim_{h \to 0} x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.10

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + lim_{h \to 0} h^{2} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.11

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + lim_{h \to 0} h^{2} \cdot lim_{h \to 0} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.12

Move the exponent $2$ from $h^{2}$ outside the limit using the Limits Power Rule.

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot lim_{h \to 0} e^{- h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.13

Move the limit into the exponent.

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{lim_{h \to 0} - h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.14

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - lim_{h \to 0} x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.15

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} - lim_{h \to 0} x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.16

Evaluate the limit of $x^{2} e^{- x}$ which is constant as $h$ approaches $0$ .

$\frac{x^{2} e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.17

Evaluate the limits by plugging in $0$ for all occurrences of $h$ .

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Step 5.1.2.17.1

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{x^{2} e^{0 - x} + 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.17.2

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{x^{2} e^{0 - x} + 2 x \cdot 0 \cdot e^{- lim_{h \to 0} h - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.17.3

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{x^{2} e^{0 - x} + 2 x \cdot 0 \cdot e^{0 - x} + {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.17.4

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{x^{2} e^{0 - x} + 2 x \cdot 0 \cdot e^{0 - x} + 0^{2} \cdot e^{- lim_{h \to 0} h - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.17.5

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{x^{2} e^{0 - x} + 2 x \cdot 0 \cdot e^{0 - x} + 0^{2} \cdot e^{0 - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18

Simplify the answer.

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Step 5.1.2.18.1

Combine the opposite terms in $x^{2} e^{0 - x} + 2 x \cdot 0 \cdot e^{0 - x} + 0^{2} \cdot e^{0 - x} - x^{2} e^{- x}$ .

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Step 5.1.2.18.1.1

Subtract $x$ from $0$ .

$\frac{x^{2} e^{- x} + 2 x \cdot 0 \cdot e^{0 - x} + 0^{2} \cdot e^{0 - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.1.2

Subtract $x$ from $0$ .

$\frac{x^{2} e^{- x} + 2 x \cdot 0 \cdot e^{- x} + 0^{2} \cdot e^{0 - x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.1.3

Subtract $x$ from $0$ .

$\frac{x^{2} e^{- x} + 2 x \cdot 0 \cdot e^{- x} + 0^{2} \cdot e^{- x} - x^{2} e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.1.4

Subtract $x^{2} e^{- x}$ from $x^{2} e^{- x}$ .

$\frac{2 x \cdot 0 \cdot e^{- x} + 0^{2} \cdot e^{- x} + 0}{lim_{h \to 0} h}$

Step 5.1.2.18.1.5

Add $2 x \cdot 0 \cdot e^{- x} + 0^{2} \cdot e^{- x}$ and $0$ .

$\frac{2 x \cdot 0 \cdot e^{- x} + 0^{2} \cdot e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.2

Simplify each term.

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Step 5.1.2.18.2.1

Multiply $2 x \cdot 0$ .

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Step 5.1.2.18.2.1.1

Multiply $0$ by $2$ .

$\frac{0 x \cdot e^{- x} + 0^{2} \cdot e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.2.1.2

Multiply $0$ by $x$ .

$\frac{0 \cdot e^{- x} + 0^{2} \cdot e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.2.2

Multiply $0$ by $e^{- x}$ .

$\frac{0 + 0^{2} \cdot e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.2.3

Raising $0$ to any positive power yields $0$ .

$\frac{0 + 0 \cdot e^{- x}}{lim_{h \to 0} h}$

Step 5.1.2.18.2.4

Multiply $0$ by $e^{- x}$ .

$\frac{0 + 0}{lim_{h \to 0} h}$

Step 5.1.2.18.3

Add $0$ and $0$ .

$\frac{0}{lim_{h \to 0} h}$

Step 5.1.3

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{0}{0}$

Step 5.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 5.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{h \to 0} \frac{x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - x^{2} e^{- x}}{h} = lim_{h \to 0} \frac{\frac{d}{d h} [x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$lim_{h \to 0} \frac{\frac{d}{d h} [x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.2

By the Sum Rule, the derivative of $x^{2} e^{- h - x} + 2 h x e^{- h - x} + h^{2} e^{- h - x} - x^{2} e^{- x}$ with respect to $h$ is $\frac{d}{d h} [x^{2} e^{- h - x}] + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [x^{2} e^{- h - x}] + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3

Evaluate $\frac{d}{d h} [x^{2} e^{- h - x}]$ .

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Step 5.3.3.1

Since $x^{2}$ is constant with respect to $h$ , the derivative of $x^{2} e^{- h - x}$ with respect to $h$ is $x^{2} \frac{d}{d h} [e^{- h - x}]$ .

$lim_{h \to 0} \frac{x^{2} \frac{d}{d h} [e^{- h - x}] + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d h} [f (g (h))]$ is $f' (g (h)) g' (h)$ where $f (h) = e^{h}$ and $g (h) = - h - x$ .

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Step 5.3.3.2.1

To apply the Chain Rule, set $u_{1}$ as $- h - x$ .

$lim_{h \to 0} \frac{x^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d h} [- h - x]) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$lim_{h \to 0} \frac{x^{2} (e^{u_{1}} \frac{d}{d h} [- h - x]) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.2.3

Replace all occurrences of $u_{1}$ with $- h - x$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} \frac{d}{d h} [- h - x]) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.3

By the Sum Rule, the derivative of $- h - x$ with respect to $h$ is $\frac{d}{d h} [- h] + \frac{d}{d h} [- x]$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} (\frac{d}{d h} [- h] + \frac{d}{d h} [- x])) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.4

Since $- 1$ is constant with respect to $h$ , the derivative of $- h$ with respect to $h$ is $- \frac{d}{d h} [h]$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} (- \frac{d}{d h} [h] + \frac{d}{d h} [- x])) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.5

Differentiate using the Power Rule which states that $\frac{d}{d h} [h^{n}]$ is $n h^{n - 1}$ where $n = 1$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} (- 1 \cdot 1 + \frac{d}{d h} [- x])) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.6

Since $- x$ is constant with respect to $h$ , the derivative of $- x$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} (- 1 \cdot 1 + 0)) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.7

Multiply $- 1$ by $1$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} (- 1 + 0)) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.8

Add $- 1$ and $0$ .

$lim_{h \to 0} \frac{x^{2} (e^{- h - x} \cdot - 1) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.9

Move $- 1$ to the left of $e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- 1 \cdot e^{- h - x}) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.10

Rewrite $- 1 e^{- h - x}$ as $- e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + \frac{d}{d h} [2 h x e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4

Evaluate $\frac{d}{d h} [2 h x e^{- h - x}]$ .

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Step 5.3.4.1

Since $2 x$ is constant with respect to $h$ , the derivative of $2 h x e^{- h - x}$ with respect to $h$ is $2 x \frac{d}{d h} [h e^{- h - x}]$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x \frac{d}{d h} [h e^{- h - x}] + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.2

Differentiate using the Product Rule which states that $\frac{d}{d h} [f (h) g (h)]$ is $f (h) \frac{d}{d h} [g (h)] + g (h) \frac{d}{d h} [f (h)]$ where $f (h) = h$ and $g (h) = e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h \frac{d}{d h} [e^{- h - x}] + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.3

Differentiate using the chain rule, which states that $\frac{d}{d h} [f (g (h))]$ is $f' (g (h)) g' (h)$ where $f (h) = e^{h}$ and $g (h) = - h - x$ .

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Step 5.3.4.3.1

To apply the Chain Rule, set $u_{2}$ as $- h - x$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d h} [- h - x]) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{u_{2}} \frac{d}{d h} [- h - x]) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.3.3

Replace all occurrences of $u_{2}$ with $- h - x$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} \frac{d}{d h} [- h - x]) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.4

By the Sum Rule, the derivative of $- h - x$ with respect to $h$ is $\frac{d}{d h} [- h] + \frac{d}{d h} [- x]$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} (\frac{d}{d h} [- h] + \frac{d}{d h} [- x])) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.5

Since $- 1$ is constant with respect to $h$ , the derivative of $- h$ with respect to $h$ is $- \frac{d}{d h} [h]$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} (- \frac{d}{d h} [h] + \frac{d}{d h} [- x])) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.6

Differentiate using the Power Rule which states that $\frac{d}{d h} [h^{n}]$ is $n h^{n - 1}$ where $n = 1$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} (- 1 \cdot 1 + \frac{d}{d h} [- x])) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.7

Since $- x$ is constant with respect to $h$ , the derivative of $- x$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} (- 1 \cdot 1 + 0)) + e^{- h - x} \frac{d}{d h} [h]) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.8

Differentiate using the Power Rule which states that $\frac{d}{d h} [h^{n}]$ is $n h^{n - 1}$ where $n = 1$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} (- 1 \cdot 1 + 0)) + e^{- h - x} \cdot 1) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.9

Multiply $- 1$ by $1$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} (- 1 + 0)) + e^{- h - x} \cdot 1) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.10

Add $- 1$ and $0$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (e^{- h - x} \cdot - 1) + e^{- h - x} \cdot 1) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.11

Move $- 1$ to the left of $e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- 1 \cdot e^{- h - x}) + e^{- h - x} \cdot 1) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.12

Rewrite $- 1 e^{- h - x}$ as $- e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x} \cdot 1) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.4.13

Multiply $e^{- h - x}$ by $1$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + \frac{d}{d h} [h^{2} e^{- h - x}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5

Evaluate $\frac{d}{d h} [h^{2} e^{- h - x}]$ .

Tap for more steps...

Step 5.3.5.1

Differentiate using the Product Rule which states that $\frac{d}{d h} [f (h) g (h)]$ is $f (h) \frac{d}{d h} [g (h)] + g (h) \frac{d}{d h} [f (h)]$ where $f (h) = h^{2}$ and $g (h) = e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} \frac{d}{d h} [e^{- h - x}] + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.2

Differentiate using the chain rule, which states that $\frac{d}{d h} [f (g (h))]$ is $f' (g (h)) g' (h)$ where $f (h) = e^{h}$ and $g (h) = - h - x$ .

Tap for more steps...

Step 5.3.5.2.1

To apply the Chain Rule, set $u_{3}$ as $- h - x$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (\frac{d}{d u_{3}} [e^{u_{3}}] \frac{d}{d h} [- h - x]) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{u_{3}} \frac{d}{d h} [- h - x]) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.2.3

Replace all occurrences of $u_{3}$ with $- h - x$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} \frac{d}{d h} [- h - x]) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.3

By the Sum Rule, the derivative of $- h - x$ with respect to $h$ is $\frac{d}{d h} [- h] + \frac{d}{d h} [- x]$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} (\frac{d}{d h} [- h] + \frac{d}{d h} [- x])) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.4

Since $- 1$ is constant with respect to $h$ , the derivative of $- h$ with respect to $h$ is $- \frac{d}{d h} [h]$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} (- \frac{d}{d h} [h] + \frac{d}{d h} [- x])) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.5

Differentiate using the Power Rule which states that $\frac{d}{d h} [h^{n}]$ is $n h^{n - 1}$ where $n = 1$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} (- 1 \cdot 1 + \frac{d}{d h} [- x])) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.6

Since $- x$ is constant with respect to $h$ , the derivative of $- x$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} (- 1 \cdot 1 + 0)) + e^{- h - x} \frac{d}{d h} [h^{2}] + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.7

Differentiate using the Power Rule which states that $\frac{d}{d h} [h^{n}]$ is $n h^{n - 1}$ where $n = 2$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} (- 1 \cdot 1 + 0)) + e^{- h - x} (2 h) + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.8

Multiply $- 1$ by $1$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} (- 1 + 0)) + e^{- h - x} (2 h) + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.9

Add $- 1$ and $0$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (e^{- h - x} \cdot - 1) + e^{- h - x} (2 h) + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.10

Move $- 1$ to the left of $e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (- 1 \cdot e^{- h - x}) + e^{- h - x} (2 h) + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.5.11

Rewrite $- 1 e^{- h - x}$ as $- e^{- h - x}$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (- e^{- h - x}) + e^{- h - x} (2 h) + \frac{d}{d h} [- x^{2} e^{- x}]}{\frac{d}{d h} [h]}$

Step 5.3.6

Since $- x^{2} e^{- x}$ is constant with respect to $h$ , the derivative of $- x^{2} e^{- x}$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x}) + e^{- h - x}) + h^{2} (- e^{- h - x}) + e^{- h - x} (2 h) + 0}{\frac{d}{d h} [h]}$

Step 5.3.7

Simplify.

Tap for more steps...

Step 5.3.7.1

Apply the distributive property.

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) + 2 x (h (- e^{- h - x})) + 2 x e^{- h - x} + h^{2} (- e^{- h - x}) + e^{- h - x} (2 h) + 0}{\frac{d}{d h} [h]}$

Step 5.3.7.2

Combine terms.

Tap for more steps...

Step 5.3.7.2.1

Multiply $- 1$ by $2$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) - 2 x (h (e^{- h - x})) + 2 x e^{- h - x} + h^{2} (- e^{- h - x}) + e^{- h - x} (2 h) + 0}{\frac{d}{d h} [h]}$

Step 5.3.7.2.2

Add $x^{2} (- e^{- h - x}) - 2 x (h (e^{- h - x})) + 2 x e^{- h - x} + h^{2} (- e^{- h - x}) + e^{- h - x} (2 h)$ and $0$ .

$lim_{h \to 0} \frac{x^{2} (- e^{- h - x}) - 2 x (h (e^{- h - x})) + 2 x e^{- h - x} + h^{2} (- e^{- h - x}) + e^{- h - x} (2 h)}{\frac{d}{d h} [h]}$

Step 5.3.7.3

Reorder terms.

$lim_{h \to 0} \frac{- e^{- h - x} x^{2} - 2 e^{- h - x} h x + 2 e^{- h - x} x - e^{- h - x} h^{2} + 2 e^{- h - x} h}{\frac{d}{d h} [h]}$

Step 5.3.7.4

Reorder factors in $- e^{- h - x} x^{2} - 2 e^{- h - x} h x + 2 e^{- h - x} x - e^{- h - x} h^{2} + 2 e^{- h - x} h$ .

$lim_{h \to 0} \frac{- x^{2} e^{- h - x} - 2 h x e^{- h - x} + 2 x e^{- h - x} - h^{2} e^{- h - x} + 2 h e^{- h - x}}{\frac{d}{d h} [h]}$

Step 5.3.8

Differentiate using the Power Rule which states that $\frac{d}{d h} [h^{n}]$ is $n h^{n - 1}$ where $n = 1$ .

$lim_{h \to 0} \frac{- x^{2} e^{- h - x} - 2 h x e^{- h - x} + 2 x e^{- h - x} - h^{2} e^{- h - x} + 2 h e^{- h - x}}{1}$

Step 5.4

Divide $- x^{2} e^{- h - x} - 2 h x e^{- h - x} + 2 x e^{- h - x} - h^{2} e^{- h - x} + 2 h e^{- h - x}$ by $1$ .

$lim_{h \to 0} - x^{2} e^{- h - x} - 2 h x e^{- h - x} + 2 x e^{- h - x} - h^{2} e^{- h - x} + 2 h e^{- h - x}$

Step 6

Evaluate the limit.

Tap for more steps...

Step 6.1

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$- lim_{h \to 0} x^{2} e^{- h - x} - lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.2

Move the term $x^{2}$ outside of the limit because it is constant with respect to $h$ .

$- x^{2} lim_{h \to 0} e^{- h - x} - lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.3

Move the limit into the exponent.

$- x^{2} e^{lim_{h \to 0} - h - x} - lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.4

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - lim_{h \to 0} x} - lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.5

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - lim_{h \to 0} 2 h x e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.6

Move the term $2 x$ outside of the limit because it is constant with respect to $h$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.7

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot lim_{h \to 0} e^{- h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.8

Move the limit into the exponent.

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{lim_{h \to 0} - h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.9

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - lim_{h \to 0} x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.10

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + lim_{h \to 0} 2 x e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.11

Move the term $2 x$ outside of the limit because it is constant with respect to $h$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x lim_{h \to 0} e^{- h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.12

Move the limit into the exponent.

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{lim_{h \to 0} - h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.13

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - lim_{h \to 0} x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.14

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - lim_{h \to 0} h^{2} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.15

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - lim_{h \to 0} h^{2} \cdot lim_{h \to 0} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.16

Move the exponent $2$ from $h^{2}$ outside the limit using the Limits Power Rule.

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot lim_{h \to 0} e^{- h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.17

Move the limit into the exponent.

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{lim_{h \to 0} - h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.18

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - lim_{h \to 0} x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.19

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} + lim_{h \to 0} 2 h e^{- h - x}$

Step 6.20

Move the term $2$ outside of the limit because it is constant with respect to $h$ .

$- x^{2} e^{- lim_{h \to 0} h - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} + 2 lim_{h \to 0} h e^{- h - x}$

Step 6.21

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

Step 6.22

Move the limit into the exponent.

Step 6.23

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

Step 6.24

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

Step 7

Evaluate the limits by plugging in $0$ for all occurrences of $h$ .

Tap for more steps...

Step 7.1

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} + 2 lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x}$

Step 7.2

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{- lim_{h \to 0} h - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} + 2 lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x}$

Step 7.3

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{- lim_{h \to 0} h - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} + 2 lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x}$

Step 7.4

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - {(lim_{h \to 0} h)}^{2} \cdot e^{- lim_{h \to 0} h - x} + 2 lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x}$

Step 7.5

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - 0^{2} \cdot e^{- lim_{h \to 0} h - x} + 2 lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x}$

Step 7.6

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - 0^{2} \cdot e^{0 - x} + 2 lim_{h \to 0} h \cdot e^{- lim_{h \to 0} h - x}$

Step 7.7

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - 0^{2} \cdot e^{0 - x} + 2 \cdot 0 \cdot e^{- lim_{h \to 0} h - x}$

Step 7.8

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - 0^{2} \cdot e^{0 - x} + 2 \cdot 0 \cdot e^{0 - x}$

Step 8

Simplify the answer.

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Step 8.1

Combine the opposite terms in $- x^{2} e^{0 - x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - 0^{2} \cdot e^{0 - x} + 2 \cdot 0 \cdot e^{0 - x}$ .

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Step 8.1.1

Subtract $x$ from $0$ .

$- x^{2} e^{- x} - 1 \cdot 2 x \cdot 0 \cdot e^{0 - x} + 2 x e^{0 - x} - 0^{2} \cdot e^{0 - x} + 2 \cdot 0 \cdot e^{0 - x}$

Step 8.1.2

Subtract $x$ from $0$ .

$- x^{2} e^{- x} - 1 \cdot 2 x \cdot 0 \cdot e^{- x} + 2 x e^{0 - x} - 0^{2} \cdot e^{0 - x} + 2 \cdot 0 \cdot e^{0 - x}$

Step 8.1.3

Subtract $x$ from $0$ .

$- x^{2} e^{- x} - 1 \cdot 2 x \cdot 0 \cdot e^{- x} + 2 x e^{- x} - 0^{2} \cdot e^{0 - x} + 2 \cdot 0 \cdot e^{0 - x}$

Step 8.1.4

Subtract $x$ from $0$ .

$- x^{2} e^{- x} - 1 \cdot 2 x \cdot 0 \cdot e^{- x} + 2 x e^{- x} - 0^{2} \cdot e^{- x} + 2 \cdot 0 \cdot e^{0 - x}$

Step 8.1.5

Subtract $x$ from $0$ .

$- x^{2} e^{- x} - 1 \cdot 2 x \cdot 0 \cdot e^{- x} + 2 x e^{- x} - 0^{2} \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2

Simplify each term.

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Step 8.2.1

Multiply $- 1$ by $2$ .

$- x^{2} e^{- x} - 2 x \cdot 0 \cdot e^{- x} + 2 x e^{- x} - 0^{2} \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.2

Multiply $- 2 x \cdot 0$ .

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Step 8.2.2.1

Multiply $0$ by $- 2$ .

$- x^{2} e^{- x} + 0 x \cdot e^{- x} + 2 x e^{- x} - 0^{2} \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.2.2

Multiply $0$ by $x$ .

$- x^{2} e^{- x} + 0 \cdot e^{- x} + 2 x e^{- x} - 0^{2} \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.3

Multiply $0$ by $e^{- x}$ .

$- x^{2} e^{- x} + 0 + 2 x e^{- x} - 0^{2} \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.4

Raising $0$ to any positive power yields $0$ .

$- x^{2} e^{- x} + 0 + 2 x e^{- x} - 0 \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.5

Multiply $- 1$ by $0$ .

$- x^{2} e^{- x} + 0 + 2 x e^{- x} + 0 \cdot e^{- x} + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.6

Multiply $0$ by $e^{- x}$ .

$- x^{2} e^{- x} + 0 + 2 x e^{- x} + 0 + 2 \cdot 0 \cdot e^{- x}$

Step 8.2.7

Multiply $2$ by $0$ .

$- x^{2} e^{- x} + 0 + 2 x e^{- x} + 0 + 0 \cdot e^{- x}$

Step 8.2.8

Multiply $0$ by $e^{- x}$ .

$- x^{2} e^{- x} + 0 + 2 x e^{- x} + 0 + 0$

Step 8.3

Combine the opposite terms in $- x^{2} e^{- x} + 0 + 2 x e^{- x} + 0 + 0$ .

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Step 8.3.1

Add $- x^{2} e^{- x}$ and $0$ .

$- x^{2} e^{- x} + 2 x e^{- x} + 0 + 0$

Step 8.3.2

Add $- x^{2} e^{- x} + 2 x e^{- x}$ and $0$ .

$- x^{2} e^{- x} + 2 x e^{- x} + 0$

Step 8.3.3

Add $- x^{2} e^{- x} + 2 x e^{- x}$ and $0$ .

$- x^{2} e^{- x} + 2 x e^{- x}$

Step 9