Calculus Examples

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

Step 1

Combine terms.

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Step 1.1

Write $1$ as a fraction with a common denominator.

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

Step 1.2

Combine the numerators over the common denominator.

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

Step 2

Use the properties of logarithms to simplify the limit.

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Step 2.1

Rewrite ${(\frac{x^{2} + 1}{x^{2}})}^{x}$ as $e^{\ln ({(\frac{x^{2} + 1}{x^{2}})}^{x})}$ .

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

Step 2.2

Expand $\ln ({(\frac{x^{2} + 1}{x^{2}})}^{x})$ by moving $x$ outside the logarithm.

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

Step 3

Move the limit into the exponent.

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

Step 4

Rewrite $x \ln (\frac{x^{2} + 1}{x^{2}})$ as $\frac{\ln (\frac{x^{2} + 1}{x^{2}})}{\frac{1}{x}}$ .

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

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

Move the limit inside the logarithm.

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

Step 5.1.2.2

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^{2}$ .

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

Step 5.1.2.3

Evaluate the limit.

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Step 5.1.2.3.1

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

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Step 5.1.2.3.1.1

Cancel the common factor.

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

Step 5.1.2.3.1.2

Rewrite the expression.

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

Step 5.1.2.3.2

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

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Step 5.1.2.3.2.1

Cancel the common factor.

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

Step 5.1.2.3.2.2

Rewrite the expression.

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

Step 5.1.2.3.3

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

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

Step 5.1.2.3.4

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

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

Step 5.1.2.3.5

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

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

Step 5.1.2.4

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

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

Step 5.1.2.5

Evaluate the limit.

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Step 5.1.2.5.1

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

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

Step 5.1.2.5.2

Simplify the answer.

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Step 5.1.2.5.2.1

Divide $1 + 0$ by $1$ .

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

Step 5.1.2.5.2.2

Add $1$ and $0$ .

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

Step 5.1.2.5.2.3

The natural logarithm of $1$ is $0$ .

$e^{\frac{0}{lim_{x \to \infty} \frac{1}{x}}}$

Step 5.1.3

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

$e^{\frac{0}{0}}$

Step 5.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$e^{\frac{0}{0}}$

Step 5.2

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.

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

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

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Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.2.3

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

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

Step 5.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{x^{2} + 1}{x^{2}}$ .

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

Step 5.3.4

Multiply $\frac{x^{2}}{x^{2} + 1}$ by $1$ .

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

Step 5.3.5

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{2} + 1$ and $g (x) = x^{2}$ .

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

Step 5.3.6

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

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Step 5.3.6.1

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

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

Step 5.3.6.2

Multiply $2$ by $2$ .

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

Step 5.3.7

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

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

Step 5.3.8

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

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

Step 5.3.9

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

$e^{lim_{x \to \infty} \frac{\frac{x^{2}}{x^{2} + 1} \cdot \frac{x^{2} (2 x + 0) - (x^{2} + 1) \frac{d}{d x} [x^{2}]}{x^{4}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.10

Add $2 x$ and $0$ .

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

Step 5.3.11

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 5.3.11.1

Move $x$ .

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

Step 5.3.11.2

Multiply $x$ by $x^{2}$ .

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Step 5.3.11.2.1

Raise $x$ to the power of $1$ .

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

Step 5.3.11.2.2

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

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

Step 5.3.11.3

Add $1$ and $2$ .

$e^{lim_{x \to \infty} \frac{\frac{x^{2}}{x^{2} + 1} \cdot \frac{x^{3} \cdot 2 - (x^{2} + 1) \frac{d}{d x} [x^{2}]}{x^{4}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.12

Move $2$ to the left of $x^{3}$ .

$e^{lim_{x \to \infty} \frac{\frac{x^{2}}{x^{2} + 1} \cdot \frac{2 \cdot x^{3} - (x^{2} + 1) \frac{d}{d x} [x^{2}]}{x^{4}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.13

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

$e^{lim_{x \to \infty} \frac{\frac{x^{2}}{x^{2} + 1} \cdot \frac{2 x^{3} - (x^{2} + 1) \cdot 2 x}{x^{4}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.14

Multiply $2$ by $- 1$ .

$e^{lim_{x \to \infty} \frac{\frac{x^{2}}{x^{2} + 1} \cdot \frac{2 x^{3} - 2 (x^{2} + 1) x}{x^{4}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.15

Multiply $\frac{x^{2}}{x^{2} + 1}$ by $\frac{2 x^{3} - 2 (x^{2} + 1) x}{x^{4}}$ .

$e^{lim_{x \to \infty} \frac{\frac{x^{2} (2 x^{3} - 2 (x^{2} + 1) x)}{(x^{2} + 1) x^{4}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.16

Cancel the common factors.

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Step 5.3.16.1

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

$e^{lim_{x \to \infty} \frac{\frac{x^{2} (2 x^{3} - 2 (x^{2} + 1) x)}{x^{2} (x^{2} + 1) x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.16.2

Cancel the common factor.

$e^{lim_{x \to \infty} \frac{\frac{x^{2} (2 x^{3} - 2 (x^{2} + 1) x)}{x^{2} (x^{2} + 1) x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.16.3

Rewrite the expression.

$e^{lim_{x \to \infty} \frac{\frac{2 x^{3} - 2 (x^{2} + 1) x}{(x^{2} + 1) x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.17

Simplify.

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Step 5.3.17.1

Apply the distributive property.

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

Step 5.3.17.2

Apply the distributive property.

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

Step 5.3.17.3

Apply the distributive property.

$e^{lim_{x \to \infty} \frac{\frac{2 x^{3} - 2 x^{2} x - 2 \cdot 1 x}{x^{2} x^{2} + 1 x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.17.4

Simplify the numerator.

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Step 5.3.17.4.1

Simplify each term.

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Step 5.3.17.4.1.1

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 5.3.17.4.1.1.1

Move $x$ .

$e^{lim_{x \to \infty} \frac{\frac{2 x^{3} - 2 x \cdot x^{2} - 2 \cdot 1 x}{x^{2} x^{2} + 1 x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.17.4.1.1.2

Multiply $x$ by $x^{2}$ .

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Step 5.3.17.4.1.1.2.1

Raise $x$ to the power of $1$ .

$e^{lim_{x \to \infty} \frac{\frac{2 x^{3} - 2 x^{1} x^{2} - 2 \cdot 1 x}{x^{2} x^{2} + 1 x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.17.4.1.1.2.2

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

$e^{lim_{x \to \infty} \frac{\frac{2 x^{3} - 2 x^{1 + 2} - 2 \cdot 1 x}{x^{2} x^{2} + 1 x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.17.4.1.1.3

Add $1$ and $2$ .

$e^{lim_{x \to \infty} \frac{\frac{2 x^{3} - 2 x^{3} - 2 \cdot 1 x}{x^{2} x^{2} + 1 x^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 5.3.17.4.1.2

Multiply $- 2$ by $1$ .

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

Step 5.3.17.4.2

Subtract $2 x^{3}$ from $2 x^{3}$ .

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

Step 5.3.17.4.3

Subtract $2 x$ from $0$ .

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

Step 5.3.17.5

Combine terms.

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Step 5.3.17.5.1

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 5.3.17.5.1.1

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

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

Step 5.3.17.5.1.2

Add $2$ and $2$ .

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

Step 5.3.17.5.2

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

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

Step 5.3.17.5.3

Move the negative in front of the fraction.

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

Step 5.3.17.6

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

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Step 5.3.17.6.1

Factor $x^{2}$ out of $x^{4}$ .

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

Step 5.3.17.6.2

Multiply by $1$ .

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

Step 5.3.17.6.3

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} \cdot 1$ .

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

Step 5.3.17.7

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

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Step 5.3.17.7.1

Factor $x$ out of $2 x$ .

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

Step 5.3.17.7.2

Cancel the common factors.

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Step 5.3.17.7.2.1

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

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

Step 5.3.17.7.2.2

Cancel the common factor.

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

Step 5.3.17.7.2.3

Rewrite the expression.

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

Step 5.3.18

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 5.3.19

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

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

Step 5.3.20

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

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

Step 5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.5

Combine factors.

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Step 5.5.1

Multiply $- 1$ by $- 1$ .

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

Step 5.5.2

Multiply $\frac{2}{x (x^{2} + 1)}$ by $1$ .

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

Step 5.5.3

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

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

Step 5.6

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

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Step 5.6.1

Factor $x$ out of $2 x^{2}$ .

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

Step 5.6.2

Cancel the common factors.

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Step 5.6.2.1

Cancel the common factor.

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

Step 5.6.2.2

Rewrite the expression.

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

Step 6

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

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

Step 7

Divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x^{2}$ .

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

Step 8

Evaluate the limit.

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Step 8.1

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

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Step 8.1.1

Raise $x$ to the power of $1$ .

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

Step 8.1.2

Factor $x$ out of $x^{1}$ .

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

Step 8.1.3

Cancel the common factors.

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Step 8.1.3.1

Factor $x$ out of $x^{2}$ .

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

Step 8.1.3.2

Cancel the common factor.

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

Step 8.1.3.3

Rewrite the expression.

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

Step 8.2

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

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Step 8.2.1

Cancel the common factor.

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

Step 8.2.2

Rewrite the expression.

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

Step 8.3

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

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

Step 9

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

$e^{2 \frac{0}{lim_{x \to \infty} 1 + \frac{1}{x^{2}}}}$

Step 10

Evaluate the limit.

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Step 10.1

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

$e^{2 \frac{0}{lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{1}{x^{2}}}}$

Step 10.2

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

$e^{2 \frac{0}{1 + lim_{x \to \infty} \frac{1}{x^{2}}}}$

Step 11

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x^{2}}$ approaches $0$ .

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

Step 12

Simplify terms.

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Step 12.1

Simplify the answer.

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Step 12.1.1

Add $1$ and $0$ .

$e^{2 (\frac{0}{1})}$

Step 12.1.2

Divide $0$ by $1$ .

$e^{2 \cdot 0}$

Step 12.1.3

Multiply $2$ by $0$ .

$e^{0}$

Step 12.2

Anything raised to $0$ is $1$ .

$1$