Calculus Examples

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

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

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

Step 1.1.2.2

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

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

Step 1.1.2.3

Move the limit into the exponent.

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

Step 1.1.2.4

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.1.2.5

Simplify terms.

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Step 1.1.2.5.1

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

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

Step 1.1.2.5.2

Simplify the answer.

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Step 1.1.2.5.2.1

Simplify each term.

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Step 1.1.2.5.2.1.1

Anything raised to $0$ is $1$ .

$\frac{1 - 1 \cdot 1}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.5.2.1.2

Multiply $- 1$ by $1$ .

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

Step 1.1.2.5.2.2

Subtract $1$ from $1$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

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

Step 1.1.3.1.2

Move the limit into the exponent.

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

Step 1.1.3.1.3

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

$\frac{0}{e^{lim_{x \to 0} x} - 1 \cdot 1}$

Step 1.1.3.2

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

$\frac{0}{e^{0} - 1 \cdot 1}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Anything raised to $0$ is $1$ .

$\frac{0}{1 - 1 \cdot 1}$

Step 1.1.3.3.1.2

Multiply $- 1$ by $1$ .

$\frac{0}{1 - 1}$

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 0} \frac{1 - e^{- x}}{e^{x} - 1} = lim_{x \to 0} \frac{\frac{d}{d x} [1 - e^{- x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [1 - e^{- x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.2

By the Sum Rule, the derivative of $1 - e^{- x}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- e^{- x}]$ .

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

Step 1.3.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 1.3.4

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

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Step 1.3.4.1

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

$lim_{x \to 0} \frac{0 - \frac{d}{d x} [e^{- x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.2

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

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Step 1.3.4.2.1

To apply the Chain Rule, set $u$ as $- x$ .

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

Step 1.3.4.2.2

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

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

Step 1.3.4.2.3

Replace all occurrences of $u$ with $- x$ .

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

Step 1.3.4.3

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

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

Step 1.3.4.4

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

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

Step 1.3.4.5

Multiply $- 1$ by $1$ .

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

Step 1.3.4.6

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

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

Step 1.3.4.7

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

$lim_{x \to 0} \frac{0 - - e^{- x}}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.8

Multiply $- 1$ by $- 1$ .

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

Step 1.3.4.9

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

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

Step 1.3.5

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

$lim_{x \to 0} \frac{e^{- x}}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.6

By the Sum Rule, the derivative of $e^{x} - 1$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 1]$ .

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

Step 1.3.7

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

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

Step 1.3.8

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 1.3.9

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

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

Step 1.4

Cancel the common factor of $e^{- x}$ and $e^{x}$ .

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Step 1.4.1

Factor $e^{x}$ out of $e^{- x}$ .

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

Step 1.4.2

Cancel the common factors.

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Step 1.4.2.1

Multiply by $1$ .

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

Step 1.4.2.2

Cancel the common factor.

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

Step 1.4.2.3

Rewrite the expression.

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

Step 1.4.2.4

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

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

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

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

Step 2.2

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

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

Step 3

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

$e^{- 2 \cdot 0}$

Step 4

Simplify the answer.

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Step 4.1

Multiply $- 2$ by $0$ .

$e^{0}$

Step 4.2

Anything raised to $0$ is $1$ .

$1$