Calculus Examples

$lim_{x \to \infty} \frac{\ln (x + 1)}{\log_{2} (x)}$

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.

$\frac{lim_{x \to \infty} \ln (x + 1)}{lim_{x \to \infty} \log_{2} (x)}$

Step 1.2

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

$\frac{\infty}{lim_{x \to \infty} \log_{2} (x)}$

Step 1.3

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

$\frac{\infty}{\infty}$

Step 1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 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.

$lim_{x \to \infty} \frac{\ln (x + 1)}{\log_{2} (x)} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x + 1)]}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x + 1)]}{\frac{d}{d x} [\log_{2} (x)]}$

Step 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 $g (x) = x + 1$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x + 1]}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.2.2

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

$lim_{x \to \infty} \frac{\frac{1}{u} \frac{d}{d x} [x + 1]}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.2.3

Replace all occurrences of $u$ with $x + 1$ .

$lim_{x \to \infty} \frac{\frac{1}{x + 1} \frac{d}{d x} [x + 1]}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.3

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

$lim_{x \to \infty} \frac{\frac{1}{x + 1} (\frac{d}{d x} [x] + \frac{d}{d x} [1])}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.4

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

$lim_{x \to \infty} \frac{\frac{1}{x + 1} (1 + \frac{d}{d x} [1])}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.5

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

$lim_{x \to \infty} \frac{\frac{1}{x + 1} (1 + 0)}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.6

Add $1$ and $0$ .

$lim_{x \to \infty} \frac{\frac{1}{x + 1} \cdot 1}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.7

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

$lim_{x \to \infty} \frac{\frac{1}{x + 1}}{\frac{d}{d x} [\log_{2} (x)]}$

Step 3.8

The derivative of $\log_{2} (x)$ with respect to $x$ is $\frac{1}{x \ln (2)}$ .

$lim_{x \to \infty} \frac{\frac{1}{x + 1}}{\frac{1}{x \ln (2)}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \infty} \frac{1}{x + 1} (x \ln (2))$

Step 5

Combine factors.

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Step 5.1

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

$lim_{x \to \infty} \frac{x}{x + 1} \ln (2)$

Step 5.2

Combine $\frac{x}{x + 1}$ and $\ln (2)$ .

$lim_{x \to \infty} \frac{x \ln (2)}{x + 1}$

Step 6

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

$\ln (2) lim_{x \to \infty} \frac{x}{x + 1}$

Step 7

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$ .

$\ln (2) lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{x}{x} + \frac{1}{x}}$

Step 8

Evaluate the limit.

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Step 8.1

Cancel the common factor of $x$ .

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Step 8.1.1

Cancel the common factor.

$\ln (2) lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{x}{x} + \frac{1}{x}}$

Step 8.1.2

Rewrite the expression.

$\ln (2) lim_{x \to \infty} \frac{1}{\frac{x}{x} + \frac{1}{x}}$

Step 8.2

Cancel the common factor of $x$ .

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Step 8.2.1

Cancel the common factor.

$\ln (2) lim_{x \to \infty} \frac{1}{\frac{x}{x} + \frac{1}{x}}$

Step 8.2.2

Rewrite the expression.

$\ln (2) lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}}$

Step 8.3

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

$\ln (2) \frac{lim_{x \to \infty} 1}{lim_{x \to \infty} 1 + \frac{1}{x}}$

Step 8.4

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

$\ln (2) \frac{1}{lim_{x \to \infty} 1 + \frac{1}{x}}$

Step 8.5

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

$\ln (2) \frac{1}{lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{1}{x}}$

Step 8.6

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

$\ln (2) \frac{1}{1 + lim_{x \to \infty} \frac{1}{x}}$

Step 9

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x}$ approaches $0$ .

$\ln (2) \frac{1}{1 + 0}$

Step 10

Simplify the answer.

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Step 10.1

Add $1$ and $0$ .

$\ln (2) \frac{1}{1}$

Step 10.2

Cancel the common factor of $1$ .

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Step 10.2.1

Cancel the common factor.

$\ln (2) \frac{1}{1}$

Step 10.2.2

Rewrite the expression.

$\ln (2) \cdot 1$

Step 10.3

Multiply $\ln (2)$ by $1$ .

$\ln (2)$