Calculus Examples

$lim_{x \to 0} \frac{e^{9 x} - 1 - 9 x}{x^{2}}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to 0} e^{9 x} - 1 - 9 x}{lim_{x \to 0} x^{2}}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

$\frac{lim_{x \to 0} e^{9 x} - lim_{x \to 0} 1 - lim_{x \to 0} 9 x}{lim_{x \to 0} x^{2}}$

Step 1.2.2

Move the limit into the exponent.

$\frac{e^{lim_{x \to 0} 9 x} - lim_{x \to 0} 1 - lim_{x \to 0} 9 x}{lim_{x \to 0} x^{2}}$

Step 1.2.3

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{e^{9 lim_{x \to 0} x} - lim_{x \to 0} 1 - lim_{x \to 0} 9 x}{lim_{x \to 0} x^{2}}$

Step 1.2.4

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

$\frac{e^{9 lim_{x \to 0} x} - 1 \cdot 1 - lim_{x \to 0} 9 x}{lim_{x \to 0} x^{2}}$

Step 1.2.5

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{e^{9 lim_{x \to 0} x} - 1 \cdot 1 - 9 lim_{x \to 0} x}{lim_{x \to 0} x^{2}}$

Step 1.2.6

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

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Step 1.2.6.1

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

$\frac{e^{9 \cdot 0} - 1 \cdot 1 - 9 lim_{x \to 0} x}{lim_{x \to 0} x^{2}}$

Step 1.2.6.2

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

$\frac{e^{9 \cdot 0} - 1 \cdot 1 - 9 \cdot 0}{lim_{x \to 0} x^{2}}$

Step 1.2.7

Simplify the answer.

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Step 1.2.7.1

Simplify each term.

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Step 1.2.7.1.1

Multiply $9$ by $0$ .

$\frac{e^{0} - 1 \cdot 1 - 9 \cdot 0}{lim_{x \to 0} x^{2}}$

Step 1.2.7.1.2

Anything raised to $0$ is $1$ .

$\frac{1 - 1 \cdot 1 - 9 \cdot 0}{lim_{x \to 0} x^{2}}$

Step 1.2.7.1.3

Multiply $- 1$ by $1$ .

$\frac{1 - 1 - 9 \cdot 0}{lim_{x \to 0} x^{2}}$

Step 1.2.7.1.4

Multiply $- 9$ by $0$ .

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

Step 1.2.7.2

Subtract $1$ from $1$ .

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

Step 1.2.7.3

Add $0$ and $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

$\frac{0}{{(lim_{x \to 0} x)}^{2}}$

Step 1.3.2

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

$\frac{0}{0^{2}}$

Step 1.3.3

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

$\frac{0}{0}$

Step 1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

$lim_{x \to 0} \frac{e^{9 x} - 1 - 9 x}{x^{2}} = lim_{x \to 0} \frac{\frac{d}{d x} [e^{9 x} - 1 - 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{9 x} - 1 - 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.2

By the Sum Rule, the derivative of $e^{9 x} - 1 - 9 x$ with respect to $x$ is $\frac{d}{d x} [e^{9 x}] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{9 x}] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3

Evaluate $\frac{d}{d x} [e^{9 x}]$ .

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Step 3.3.1

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 $g (x) = 9 x$ .

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Step 3.3.1.1

To apply the Chain Rule, set $u$ as $9 x$ .

$lim_{x \to 0} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d x} [9 x] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.1.2

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

$lim_{x \to 0} \frac{e^{u} \frac{d}{d x} [9 x] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.1.3

Replace all occurrences of $u$ with $9 x$ .

$lim_{x \to 0} \frac{e^{9 x} \frac{d}{d x} [9 x] + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.2

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{e^{9 x} (9 \frac{d}{d x} [x]) + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.3

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

$lim_{x \to 0} \frac{e^{9 x} (9 \cdot 1) + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.4

Multiply $9$ by $1$ .

$lim_{x \to 0} \frac{e^{9 x} \cdot 9 + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.3.5

Move $9$ to the left of $e^{9 x}$ .

$lim_{x \to 0} \frac{9 e^{9 x} + \frac{d}{d x} [- 1] + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.4

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

$lim_{x \to 0} \frac{9 e^{9 x} + 0 + \frac{d}{d x} [- 9 x]}{\frac{d}{d x} [x^{2}]}$

Step 3.5

Evaluate $\frac{d}{d x} [- 9 x]$ .

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Step 3.5.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x$ with respect to $x$ is $- 9 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{9 e^{9 x} + 0 - 9 \frac{d}{d x} [x]}{\frac{d}{d x} [x^{2}]}$

Step 3.5.2

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

$lim_{x \to 0} \frac{9 e^{9 x} + 0 - 9 \cdot 1}{\frac{d}{d x} [x^{2}]}$

Step 3.5.3

Multiply $- 9$ by $1$ .

$lim_{x \to 0} \frac{9 e^{9 x} + 0 - 9}{\frac{d}{d x} [x^{2}]}$

Step 3.6

Add $9 e^{9 x}$ and $0$ .

$lim_{x \to 0} \frac{9 e^{9 x} - 9}{\frac{d}{d x} [x^{2}]}$

Step 3.7

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

$lim_{x \to 0} \frac{9 e^{9 x} - 9}{2 x}$

Step 4

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 e^{9 x} - 9}{x}$

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.

$\frac{1}{2} \cdot \frac{lim_{x \to 0} 9 e^{9 x} - 9}{lim_{x \to 0} x}$

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

Evaluate the limit.

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Step 5.1.2.1.1

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

$\frac{1}{2} \cdot \frac{lim_{x \to 0} 9 e^{9 x} - lim_{x \to 0} 9}{lim_{x \to 0} x}$

Step 5.1.2.1.2

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot \frac{9 lim_{x \to 0} e^{9 x} - lim_{x \to 0} 9}{lim_{x \to 0} x}$

Step 5.1.2.1.3

Move the limit into the exponent.

$\frac{1}{2} \cdot \frac{9 e^{lim_{x \to 0} 9 x} - lim_{x \to 0} 9}{lim_{x \to 0} x}$

Step 5.1.2.1.4

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot \frac{9 e^{9 lim_{x \to 0} x} - lim_{x \to 0} 9}{lim_{x \to 0} x}$

Step 5.1.2.1.5

Evaluate the limit of $9$ which is constant as $x$ approaches $0$ .

$\frac{1}{2} \cdot \frac{9 e^{9 lim_{x \to 0} x} - 1 \cdot 9}{lim_{x \to 0} x}$

Step 5.1.2.2

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

$\frac{1}{2} \cdot \frac{9 e^{9 \cdot 0} - 1 \cdot 9}{lim_{x \to 0} x}$

Step 5.1.2.3

Simplify the answer.

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Step 5.1.2.3.1

Simplify each term.

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Step 5.1.2.3.1.1

Multiply $9$ by $0$ .

$\frac{1}{2} \cdot \frac{9 e^{0} - 1 \cdot 9}{lim_{x \to 0} x}$

Step 5.1.2.3.1.2

Anything raised to $0$ is $1$ .

$\frac{1}{2} \cdot \frac{9 \cdot 1 - 1 \cdot 9}{lim_{x \to 0} x}$

Step 5.1.2.3.1.3

Multiply $9$ by $1$ .

$\frac{1}{2} \cdot \frac{9 - 1 \cdot 9}{lim_{x \to 0} x}$

Step 5.1.2.3.1.4

Multiply $- 1$ by $9$ .

$\frac{1}{2} \cdot \frac{9 - 9}{lim_{x \to 0} x}$

Step 5.1.2.3.2

Subtract $9$ from $9$ .

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

Step 5.1.3

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

$\frac{1}{2} \cdot \frac{0}{0}$

Step 5.1.4

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

Undefined

$\frac{1}{2} \cdot \frac{0}{0}$

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.

$lim_{x \to 0} \frac{9 e^{9 x} - 9}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [9 e^{9 x} - 9]}{\frac{d}{d x} [x]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$\frac{1}{2} lim_{x \to 0} \frac{\frac{d}{d x} [9 e^{9 x} - 9]}{\frac{d}{d x} [x]}$

Step 5.3.2

By the Sum Rule, the derivative of $9 e^{9 x} - 9$ with respect to $x$ is $\frac{d}{d x} [9 e^{9 x}] + \frac{d}{d x} [- 9]$ .

$\frac{1}{2} lim_{x \to 0} \frac{\frac{d}{d x} [9 e^{9 x}] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3

Evaluate $\frac{d}{d x} [9 e^{9 x}]$ .

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Step 5.3.3.1

Since $9$ is constant with respect to $x$ , the derivative of $9 e^{9 x}$ with respect to $x$ is $9 \frac{d}{d x} [e^{9 x}]$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 \frac{d}{d x} [e^{9 x}] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.2

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 $g (x) = 9 x$ .

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Step 5.3.3.2.1

To apply the Chain Rule, set $u$ as $9 x$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 \frac{d}{d u} [e^{u}] \frac{d}{d x} [9 x] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.2.2

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

$\frac{1}{2} lim_{x \to 0} \frac{9 e^{u} \frac{d}{d x} [9 x] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.2.3

Replace all occurrences of $u$ with $9 x$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 e^{9 x} \frac{d}{d x} [9 x] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.3

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 e^{9 x} \cdot 9 \frac{d}{d x} [x] + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.4

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

$\frac{1}{2} lim_{x \to 0} \frac{9 e^{9 x} \cdot 9 \cdot 1 + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.5

Multiply $9$ by $1$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 e^{9 x} \cdot 9 + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.6

Move $9$ to the left of $e^{9 x}$ .

$\frac{1}{2} lim_{x \to 0} \frac{9 \cdot 9 \cdot e^{9 x} + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.3.7

Multiply $9$ by $9$ .

$\frac{1}{2} lim_{x \to 0} \frac{81 e^{9 x} + \frac{d}{d x} [- 9]}{\frac{d}{d x} [x]}$

Step 5.3.4

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

$\frac{1}{2} lim_{x \to 0} \frac{81 e^{9 x} + 0}{\frac{d}{d x} [x]}$

Step 5.3.5

Add $81 e^{9 x}$ and $0$ .

$\frac{1}{2} lim_{x \to 0} \frac{81 e^{9 x}}{\frac{d}{d x} [x]}$

Step 5.3.6

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

$\frac{1}{2} lim_{x \to 0} \frac{81 e^{9 x}}{1}$

Step 5.4

Divide $81 e^{9 x}$ by $1$ .

$\frac{1}{2} lim_{x \to 0} 81 e^{9 x}$

Step 6

Evaluate the limit.

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Step 6.1

Move the term $81$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot 81 lim_{x \to 0} e^{9 x}$

Step 6.2

Move the limit into the exponent.

$\frac{1}{2} \cdot 81 e^{lim_{x \to 0} 9 x}$

Step 6.3

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} \cdot 81 e^{9 lim_{x \to 0} x}$

Step 7

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

$\frac{1}{2} \cdot 81 e^{9 \cdot 0}$

Step 8

Simplify the answer.

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Step 8.1

Combine $\frac{1}{2}$ and $81$ .

$\frac{81}{2} e^{9 \cdot 0}$

Step 8.2

Multiply $9$ by $0$ .

$\frac{81}{2} e^{0}$

Step 8.3

Anything raised to $0$ is $1$ .

$\frac{81}{2} \cdot 1$

Step 8.4

Multiply $\frac{81}{2}$ by $1$ .

$\frac{81}{2}$