Calculus Examples

$lim_{x \to 0} \frac{8^{2 x} - 1}{8^{x} - 1}$

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.

$\frac{lim_{x \to 0} 8^{2 x} - 1}{lim_{x \to 0} 8^{x} - 1}$

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 $0$ .

$\frac{lim_{x \to 0} 8^{2 x} - lim_{x \to 0} 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.1.2

Move the limit into the exponent.

$\frac{8^{lim_{x \to 0} 2 x} - lim_{x \to 0} 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.1.3

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

$\frac{8^{2 lim_{x \to 0} x} - lim_{x \to 0} 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.1.4

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

$\frac{8^{2 lim_{x \to 0} x} - 1 \cdot 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.2

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

$\frac{8^{2 \cdot 0} - 1 \cdot 1}{lim_{x \to 0} 8^{x} - 1}$

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 $0$ .

$\frac{8^{0} - 1 \cdot 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.3.1.2

Anything raised to $0$ is $1$ .

$\frac{1 - 1 \cdot 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.3.1.3

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 0} 8^{x} - 1}$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} 8^{x} - 1}$

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 $0$ .

$\frac{0}{lim_{x \to 0} 8^{x} - lim_{x \to 0} 1}$

Step 1.1.3.1.2

Move the limit into the exponent.

$\frac{0}{8^{lim_{x \to 0} x} - lim_{x \to 0} 1}$

Step 1.1.3.1.3

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

$\frac{0}{8^{lim_{x \to 0} x} - 1 \cdot 1}$

Step 1.1.3.2

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

$\frac{0}{8^{0} - 1 \cdot 1}$

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.

$lim_{x \to 0} \frac{8^{2 x} - 1}{8^{x} - 1} = lim_{x \to 0} \frac{\frac{d}{d x} [8^{2 x} - 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [8^{2 x} - 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.2

By the Sum Rule, the derivative of $8^{2 x} - 1$ with respect to $x$ is $\frac{d}{d x} [8^{2 x}] + \frac{d}{d x} [- 1]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [8^{2 x}] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [8^{2 x}]$ .

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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 $f (x) = 8^{x}$ and $g (x) = 2 x$ .

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Step 1.3.3.1.1

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

$lim_{x \to 0} \frac{\frac{d}{d u} [8^{u}] \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

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 $a$ = $8$ .

$lim_{x \to 0} \frac{8^{u} \ln (8) \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.1.3

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

$lim_{x \to 0} \frac{8^{2 x} \ln (8) \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.2

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

$lim_{x \to 0} \frac{8^{2 x} \ln (8) (2 \frac{d}{d x} [x]) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

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 $n = 1$ .

$lim_{x \to 0} \frac{8^{2 x} \ln (8) (2 \cdot 1) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.4

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{8^{2 x} \ln (8) \cdot 2 + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.5

Rewrite $8$ as $2^{3}$ .

$lim_{x \to 0} \frac{2 \cdot {(2^{3})}^{2 x} \ln (8) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.6

Multiply the exponents in ${(2^{3})}^{2 x}$ .

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Step 1.3.3.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$lim_{x \to 0} \frac{2 \cdot 2^{3 (2 x)} \ln (8) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.6.2

Multiply $2$ by $3$ .

$lim_{x \to 0} \frac{2 \cdot 2^{6 x} \ln (8) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8) + \frac{d}{d x} [- 1]}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.4

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

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8) + 0}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.5

Add $2^{1 + 6 x} \ln (8)$ and $0$ .

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8)}{\frac{d}{d x} [8^{x} - 1]}$

Step 1.3.6

By the Sum Rule, the derivative of $8^{x} - 1$ with respect to $x$ is $\frac{d}{d x} [8^{x}] + \frac{d}{d x} [- 1]$ .

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8)}{\frac{d}{d x} [8^{x}] + \frac{d}{d x} [- 1]}$

Step 1.3.7

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $8$ .

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8)}{8^{x} \ln (8) + \frac{d}{d x} [- 1]}$

Step 1.3.8

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

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8)}{8^{x} \ln (8) + 0}$

Step 1.3.9

Add $8^{x} \ln (8)$ and $0$ .

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8)}{8^{x} \ln (8)}$

Step 1.4

Cancel the common factor of $\ln (8)$ .

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Step 1.4.1

Cancel the common factor.

$lim_{x \to 0} \frac{2^{1 + 6 x} \ln (8)}{8^{x} \ln (8)}$

Step 1.4.2

Rewrite the expression.

$lim_{x \to 0} \frac{2^{1 + 6 x}}{8^{x}}$

Step 2

Evaluate the limit.

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Step 2.1

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

$\frac{lim_{x \to 0} 2^{1 + 6 x}}{lim_{x \to 0} 8^{x}}$

Step 2.2

Move the limit into the exponent.

$\frac{2^{lim_{x \to 0} 1 + 6 x}}{lim_{x \to 0} 8^{x}}$

Step 2.3

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

$\frac{2^{lim_{x \to 0} 1 + lim_{x \to 0} 6 x}}{lim_{x \to 0} 8^{x}}$

Step 2.4

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

$\frac{2^{1 + lim_{x \to 0} 6 x}}{lim_{x \to 0} 8^{x}}$

Step 2.5

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$\frac{2^{1 + 6 lim_{x \to 0} x}}{lim_{x \to 0} 8^{x}}$

Step 2.6

Move the limit into the exponent.

$\frac{2^{1 + 6 lim_{x \to 0} x}}{8^{lim_{x \to 0} x}}$

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 $x$ by plugging in $0$ for $x$ .

$\frac{2^{1 + 6 \cdot 0}}{8^{lim_{x \to 0} x}}$

Step 3.2

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

$\frac{2^{1 + 6 \cdot 0}}{8^{0}}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $6$ by $0$ .

$\frac{2^{1 + 0}}{8^{0}}$

Step 4.1.2

Add $1$ and $0$ .

$\frac{2^{1}}{8^{0}}$

Step 4.1.3

Evaluate the exponent.

$\frac{2}{8^{0}}$

Step 4.2

Anything raised to $0$ is $1$ .

$\frac{2}{1}$

Step 4.3

Divide $2$ by $1$ .

$2$