Calculus Examples

$lim_{x \to \infty} \frac{4 x}{\ln (x^{2} + 1)}$

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.

$\frac{lim_{x \to \infty} 4 x}{lim_{x \to \infty} \ln (x^{2} + 1)}$

Step 1.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

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.

$lim_{x \to \infty} \frac{4 x}{\ln (x^{2} + 1)} = lim_{x \to \infty} \frac{\frac{d}{d x} [4 x]}{\frac{d}{d x} [\ln (x^{2} + 1)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [4 x]}{\frac{d}{d x} [\ln (x^{2} + 1)]}$

Step 3.2

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

$lim_{x \to \infty} \frac{4 \frac{d}{d x} [x]}{\frac{d}{d x} [\ln (x^{2} + 1)]}$

Step 3.3

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

$lim_{x \to \infty} \frac{4 \cdot 1}{\frac{d}{d x} [\ln (x^{2} + 1)]}$

Step 3.4

Multiply $4$ by $1$ .

$lim_{x \to \infty} \frac{4}{\frac{d}{d x} [\ln (x^{2} + 1)]}$

Step 3.5

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 $g (x) = x^{2} + 1$ .

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Step 3.5.1

To apply the Chain Rule, set $u$ as $x^{2} + 1$ .

$lim_{x \to \infty} \frac{4}{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} + 1]}$

Step 3.5.2

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

$lim_{x \to \infty} \frac{4}{\frac{1}{u} \frac{d}{d x} [x^{2} + 1]}$

Step 3.5.3

Replace all occurrences of $u$ with $x^{2} + 1$ .

$lim_{x \to \infty} \frac{4}{\frac{1}{x^{2} + 1} \frac{d}{d x} [x^{2} + 1]}$

Step 3.6

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

$lim_{x \to \infty} \frac{4}{\frac{1}{x^{2} + 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1])}$

Step 3.7

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

$lim_{x \to \infty} \frac{4}{\frac{1}{x^{2} + 1} (2 x + \frac{d}{d x} [1])}$

Step 3.8

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

$lim_{x \to \infty} \frac{4}{\frac{1}{x^{2} + 1} (2 x + 0)}$

Step 3.9

Add $2 x$ and $0$ .

$lim_{x \to \infty} \frac{4}{\frac{1}{x^{2} + 1} (2 x)}$

Step 3.10

Combine $2$ and $\frac{1}{x^{2} + 1}$ .

$lim_{x \to \infty} \frac{4}{\frac{2}{x^{2} + 1} x}$

Step 3.11

Combine $\frac{2}{x^{2} + 1}$ and $x$ .

$lim_{x \to \infty} \frac{4}{\frac{2 x}{x^{2} + 1}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \infty} 4 \frac{x^{2} + 1}{2 x}$

Step 5

Simplify terms.

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Step 5.1

Combine $4$ and $\frac{x^{2} + 1}{2 x}$ .

$lim_{x \to \infty} \frac{4 (x^{2} + 1)}{2 x}$

Step 5.2

Cancel the common factor of $4$ and $2$ .

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Step 5.2.1

Factor $2$ out of $4 (x^{2} + 1)$ .

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

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $2$ out of $2 x$ .

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

Step 5.2.2.2

Cancel the common factor.

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

Step 5.2.2.3

Rewrite the expression.

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

Step 6

Apply L'Hospital's rule.

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Step 6.1

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

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Step 6.1.1

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

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

Step 6.1.2

Evaluate the limit of the numerator.

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Step 6.1.2.1

Simplify by multiplying through.

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Step 6.1.2.1.1

Apply the distributive property.

$\frac{lim_{x \to \infty} 2 x^{2} + 2 \cdot 1}{lim_{x \to \infty} x}$

Step 6.1.2.1.2

Multiply $2$ by $1$ .

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

Step 6.1.2.2

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\frac{\infty}{lim_{x \to \infty} x}$

Step 6.1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

Step 6.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 6.2

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

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

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

Step 6.3.2

Since $2$ is constant with respect to $x$ , the derivative of $2 (x^{2} + 1)$ with respect to $x$ is $2 \frac{d}{d x} [x^{2} + 1]$ .

$lim_{x \to \infty} \frac{2 \frac{d}{d x} [x^{2} + 1]}{\frac{d}{d x} [x]}$

Step 6.3.3

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

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

Step 6.3.4

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

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

Step 6.3.5

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

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

Step 6.3.6

Add $2 x$ and $0$ .

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

Step 6.3.7

Multiply $2$ by $2$ .

$lim_{x \to \infty} \frac{4 x}{\frac{d}{d x} [x]}$

Step 6.3.8

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

$lim_{x \to \infty} \frac{4 x}{1}$

Step 6.4

Divide $4 x$ by $1$ .

$lim_{x \to \infty} 4 x$

Step 7

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$\infty$