Calculus Examples

$lim_{x \to 0} \frac{3^{2 x} - 1}{3^{x + 2} - 9}$

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.

$\frac{lim_{x \to 0} 3^{2 x} - 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

$\frac{lim_{x \to 0} 3^{2 x} - lim_{x \to 0} 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.1.2

Move the limit into the exponent.

$\frac{3^{lim_{x \to 0} 2 x} - lim_{x \to 0} 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.1.3

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

$\frac{3^{2 lim_{x \to 0} x} - lim_{x \to 0} 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.1.4

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

$\frac{3^{2 lim_{x \to 0} x} - 1 \cdot 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.2

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

$\frac{3^{2 \cdot 0} - 1 \cdot 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Multiply $2$ by $0$ .

$\frac{3^{0} - 1 \cdot 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.3.1.2

Anything raised to $0$ is $1$ .

$\frac{1 - 1 \cdot 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.3.1.3

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 0} 3^{x + 2} - 9}$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} 3^{x + 2} - 9}$

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 $0$ .

$\frac{0}{lim_{x \to 0} 3^{x + 2} - lim_{x \to 0} 9}$

Step 1.1.3.1.2

Move the limit into the exponent.

$\frac{0}{3^{lim_{x \to 0} x + 2} - lim_{x \to 0} 9}$

Step 1.1.3.1.3

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

$\frac{0}{3^{lim_{x \to 0} x + lim_{x \to 0} 2} - lim_{x \to 0} 9}$

Step 1.1.3.1.4

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

$\frac{0}{3^{lim_{x \to 0} x + 2} - lim_{x \to 0} 9}$

Step 1.1.3.1.5

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

$\frac{0}{3^{lim_{x \to 0} x + 2} - 1 \cdot 9}$

Step 1.1.3.2

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

$\frac{0}{3^{0 + 2} - 1 \cdot 9}$

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

Add $0$ and $2$ .

$\frac{0}{3^{2} - 1 \cdot 9}$

Step 1.1.3.3.1.2

Raise $3$ to the power of $2$ .

$\frac{0}{9 - 1 \cdot 9}$

Step 1.1.3.3.1.3

Multiply $- 1$ by $9$ .

$\frac{0}{9 - 9}$

Step 1.1.3.3.2

Subtract $9$ from $9$ .

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{3^{2 x} - 1}{3^{x + 2} - 9} = lim_{x \to 0} \frac{\frac{d}{d x} [3^{2 x} - 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [3^{2 x} - 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.2

By the Sum Rule, the derivative of $3^{2 x} - 1$ with respect to $x$ is $\frac{d}{d x} [3^{2 x}] + \frac{d}{d x} [- 1]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [3^{2 x}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [3^{2 x}]$ .

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Step 1.3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 3^{x}$ and $g (x) = 2 x$ .

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Step 1.3.3.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to 0} \frac{\frac{d}{d u} [3^{u}] \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3.1.2

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

$lim_{x \to 0} \frac{3^{u} \ln (3) \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3.1.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to 0} \frac{3^{2 x} \ln (3) \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3.2

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

$lim_{x \to 0} \frac{3^{2 x} \ln (3) (2 \frac{d}{d x} [x]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3.3

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

$lim_{x \to 0} \frac{3^{2 x} \ln (3) (2 \cdot 1) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3.4

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{3^{2 x} \ln (3) \cdot 2 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.3.5

Move $2$ to the left of $3^{2 x} \ln (3)$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.4

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

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3) + 0}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.5

Add $2 \cdot 3^{2 x} \ln (3)$ and $0$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{\frac{d}{d x} [3^{x + 2} - 9]}$

Step 1.3.6

By the Sum Rule, the derivative of $3^{x + 2} - 9$ with respect to $x$ is $\frac{d}{d x} [3^{x + 2}] + \frac{d}{d x} [- 9]$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{\frac{d}{d x} [3^{x + 2}] + \frac{d}{d x} [- 9]}$

Step 1.3.7

Evaluate $\frac{d}{d x} [3^{x + 2}]$ .

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Step 1.3.7.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 3^{x}$ and $g (x) = x + 2$ .

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Step 1.3.7.1.1

To apply the Chain Rule, set $u$ as $x + 2$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{\frac{d}{d u} [3^{u}] \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 9]}$

Step 1.3.7.1.2

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

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{u} \ln (3) \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 9]}$

Step 1.3.7.1.3

Replace all occurrences of $u$ with $x + 2$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) \frac{d}{d x} [x + 2] + \frac{d}{d x} [- 9]}$

Step 1.3.7.2

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

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) (\frac{d}{d x} [x] + \frac{d}{d x} [2]) + \frac{d}{d x} [- 9]}$

Step 1.3.7.3

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

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) (1 + \frac{d}{d x} [2]) + \frac{d}{d x} [- 9]}$

Step 1.3.7.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) (1 + 0) + \frac{d}{d x} [- 9]}$

Step 1.3.7.5

Add $1$ and $0$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) \cdot 1 + \frac{d}{d x} [- 9]}$

Step 1.3.7.6

Multiply $3^{x + 2}$ by $1$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) + \frac{d}{d x} [- 9]}$

Step 1.3.8

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

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3) + 0}$

Step 1.3.9

Add $3^{x + 2} \ln (3)$ and $0$ .

$lim_{x \to 0} \frac{2 \cdot 3^{2 x} \ln (3)}{3^{x + 2} \ln (3)}$

Step 1.4

Reduce.

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Step 1.4.1

Cancel the common factor of $3^{2 x}$ and $3^{x + 2}$ .

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Step 1.4.1.1

Factor $3^{x + 2}$ out of $2 \cdot 3^{2 x} \ln (3)$ .

$lim_{x \to 0} \frac{3^{x + 2} (2 \cdot 3^{2 x - (x + 2)} \ln (3))}{3^{x + 2} \ln (3)}$

Step 1.4.1.2

Cancel the common factors.

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Step 1.4.1.2.1

Factor $3^{x + 2}$ out of $3^{x + 2} \ln (3)$ .

$lim_{x \to 0} \frac{3^{x + 2} (2 \cdot 3^{2 x - (x + 2)} \ln (3))}{3^{x + 2} (\ln (3))}$

Step 1.4.1.2.2

Cancel the common factor.

$lim_{x \to 0} \frac{3^{x + 2} (2 \cdot 3^{2 x - (x + 2)} \ln (3))}{3^{x + 2} \ln (3)}$

Step 1.4.1.2.3

Rewrite the expression.

$lim_{x \to 0} \frac{2 \cdot 3^{2 x - (x + 2)} \ln (3)}{\ln (3)}$

Step 1.4.2

Cancel the common factor of $\ln (3)$ .

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Step 1.4.2.1

Cancel the common factor.

$lim_{x \to 0} \frac{2 \cdot 3^{2 x - (x + 2)} \ln (3)}{\ln (3)}$

Step 1.4.2.2

Divide $2 \cdot 3^{2 x - (x + 2)}$ by $1$ .

$lim_{x \to 0} 2 \cdot 3^{2 x - (x + 2)}$

Step 2

Evaluate the limit.

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Step 2.1

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

$2 lim_{x \to 0} 3^{2 x - (x + 2)}$

Step 2.2

Move the limit into the exponent.

$2 \cdot 3^{lim_{x \to 0} 2 x - (x + 2)}$

Step 3

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

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Step 3.1

Evaluate the limit of $2 x - (x + 2)$ by plugging in $0$ for $x$ .

$2 \cdot 3^{2 \cdot 0 - (0 + 2)}$

Step 3.2

Simplify each term.

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Step 3.2.1

Multiply $2$ by $0$ .

$2 \cdot 3^{0 - (0 + 2)}$

Step 3.2.2

Add $0$ and $2$ .

$2 \cdot 3^{0 - 1 \cdot 2}$

Step 3.2.3

Multiply $- 1$ by $2$ .

$2 \cdot 3^{0 - 2}$

Step 3.3

Subtract $2$ from $0$ .

$2 \cdot 3^{- 2}$

Step 4

Simplify the answer.

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Step 4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 \cdot \frac{1}{3^{2}}$

Step 4.2

Raise $3$ to the power of $2$ .

$2 \cdot \frac{1}{9}$

Step 4.3

Combine $2$ and $\frac{1}{9}$ .

$\frac{2}{9}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{2}{9}$

Decimal Form:

$0.\overline{2}$