Calculus Examples

$lim_{x \to \infty} {(17 x)}^{\frac{\ln (6) + 1}{\ln (4 x) + 1}}$

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(17 x)}^{\frac{\ln (6) + 1}{\ln (4 x) + 1}}$ as $e^{\ln ({(17 x)}^{\frac{\ln (6) + 1}{\ln (4 x) + 1}})}$ .

$lim_{x \to \infty} e^{\ln ({(17 x)}^{\frac{\ln (6) + 1}{\ln (4 x) + 1}})}$

Step 1.2

Expand $\ln ({(17 x)}^{\frac{\ln (6) + 1}{\ln (4 x) + 1}})$ by moving $\frac{\ln (6) + 1}{\ln (4 x) + 1}$ outside the logarithm.

$lim_{x \to \infty} e^{\frac{\ln (6) + 1}{\ln (4 x) + 1} \ln (17 x)}$

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

$e^{lim_{x \to \infty} \frac{\ln (6) + 1}{\ln (4 x) + 1} \ln (17 x)}$

Step 2.2

Combine $\frac{\ln (6) + 1}{\ln (4 x) + 1}$ and $\ln (17 x)$ .

$e^{lim_{x \to \infty} \frac{(\ln (6) + 1) \ln (17 x)}{\ln (4 x) + 1}}$

Step 2.3

Move the term $\ln (6) + 1$ outside of the limit because it is constant with respect to $x$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\ln (17 x)}{\ln (4 x) + 1}}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

$e^{(\ln (6) + 1) \frac{lim_{x \to \infty} \ln (17 x)}{lim_{x \to \infty} \ln (4 x) + 1}}$

Step 3.1.2

As log approaches infinity, the value goes to $\infty$ .

$e^{(\ln (6) + 1) \frac{\infty}{lim_{x \to \infty} \ln (4 x) + 1}}$

Step 3.1.3

Evaluate the limit of the denominator.

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Step 3.1.3.1

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

$e^{(\ln (6) + 1) \frac{\infty}{lim_{x \to \infty} \ln (4 x) + lim_{x \to \infty} 1}}$

Step 3.1.3.2

As log approaches infinity, the value goes to $\infty$ .

$e^{(\ln (6) + 1) \frac{\infty}{\infty + lim_{x \to \infty} 1}}$

Step 3.1.3.3

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

$e^{(\ln (6) + 1) \frac{\infty}{\infty + 1}}$

Step 3.1.3.4

Infinity plus or minus a number is infinity.

$e^{(\ln (6) + 1) \frac{\infty}{\infty}}$

Step 3.1.3.5

Infinity divided by infinity is undefined.

Undefined

$e^{(\ln (6) + 1) \frac{\infty}{\infty}}$

Step 3.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{(\ln (6) + 1) \frac{\infty}{\infty}}$

Step 3.2

Since $\frac{\infty}{\infty}$ 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.

$lim_{x \to \infty} \frac{\ln (17 x)}{\ln (4 x) + 1} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (17 x)]}{\frac{d}{d x} [\ln (4 x) + 1]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (17 x)]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 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 $f (x) = \ln (x)$ and $g (x) = 17 x$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u$ as $17 x$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [17 x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{u} \frac{d}{d x} [17 x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.2.3

Replace all occurrences of $u$ with $17 x$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{17 x} \frac{d}{d x} [17 x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.3

Since $17$ is constant with respect to $x$ , the derivative of $17 x$ with respect to $x$ is $17 \frac{d}{d x} [x]$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{17 x} \cdot 17 \frac{d}{d x} [x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.4

Combine $17$ and $\frac{1}{17 x}$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{17}{17 x} \frac{d}{d x} [x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.5

Cancel the common factor of $17$ .

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Step 3.3.5.1

Cancel the common factor.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{17}{17 x} \frac{d}{d x} [x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.5.2

Rewrite the expression.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x} \frac{d}{d x} [x]}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.6

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

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x} \cdot 1}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.7

Multiply $\frac{1}{x}$ by $1$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{d}{d x} [\ln (4 x) + 1]}}$

Step 3.3.8

By the Sum Rule, the derivative of $\ln (4 x) + 1$ with respect to $x$ is $\frac{d}{d x} [\ln (4 x)] + \frac{d}{d x} [1]$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{d}{d x} [\ln (4 x)] + \frac{d}{d x} [1]}}$

Step 3.3.9

Evaluate $\frac{d}{d x} [\ln (4 x)]$ .

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Step 3.3.9.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 4 x$ .

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Step 3.3.9.1.1

To apply the Chain Rule, set $u$ as $4 x$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [4 x] + \frac{d}{d x} [1]}}$

Step 3.3.9.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{u} \frac{d}{d x} [4 x] + \frac{d}{d x} [1]}}$

Step 3.3.9.1.3

Replace all occurrences of $u$ with $4 x$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [1]}}$

Step 3.3.9.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{4 x} \cdot 4 \frac{d}{d x} [x] + \frac{d}{d x} [1]}}$

Step 3.3.9.3

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

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{4 x} \cdot 4 \cdot 1 + \frac{d}{d x} [1]}}$

Step 3.3.9.4

Multiply $4$ by $1$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{4 x} \cdot 4 + \frac{d}{d x} [1]}}$

Step 3.3.9.5

Combine $\frac{1}{4 x}$ and $4$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{4}{4 x} + \frac{d}{d x} [1]}}$

Step 3.3.9.6

Cancel the common factor of $4$ .

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Step 3.3.9.6.1

Cancel the common factor.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{4}{4 x} + \frac{d}{d x} [1]}}$

Step 3.3.9.6.2

Rewrite the expression.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{x} + \frac{d}{d x} [1]}}$

Step 3.3.10

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

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

Step 3.3.11

Add $\frac{1}{x}$ and $0$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{\frac{1}{x}}{\frac{1}{x}}}$

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{1}{x} x}$

Step 3.5

Combine $\frac{1}{x}$ and $x$ .

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{x}{x}}$

Step 3.6

Cancel the common factor of $x$ .

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Step 3.6.1

Cancel the common factor.

$e^{(\ln (6) + 1) lim_{x \to \infty} \frac{x}{x}}$

Step 3.6.2

Rewrite the expression.

$e^{(\ln (6) + 1) lim_{x \to \infty} 1}$

Step 4

Evaluate the limit.

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Step 4.1

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

$e^{(\ln (6) + 1) \cdot 1}$

Step 4.2

Multiply $\ln (6) + 1$ by $1$ .

$e^{\ln (6) + 1}$