Calculus Examples

$lim_{x \to \infty} {(1 + \frac{1}{x})}^{4 x + 7}$

Step 1

Combine terms.

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Step 1.1

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

$lim_{x \to \infty} {(\frac{x}{x} + \frac{1}{x})}^{4 x + 7}$

Step 1.2

Combine the numerators over the common denominator.

$lim_{x \to \infty} {(\frac{x + 1}{x})}^{4 x + 7}$

Step 2

Use the properties of logarithms to simplify the limit.

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Step 2.1

Rewrite ${(\frac{x + 1}{x})}^{4 x + 7}$ as $e^{\ln ({(\frac{x + 1}{x})}^{4 x + 7})}$ .

$lim_{x \to \infty} e^{\ln ({(\frac{x + 1}{x})}^{4 x + 7})}$

Step 2.2

Expand $\ln ({(\frac{x + 1}{x})}^{4 x + 7})$ by moving $4 x + 7$ outside the logarithm.

$lim_{x \to \infty} e^{(4 x + 7) \ln (\frac{x + 1}{x})}$

Step 3

Move the limit into the exponent.

$e^{lim_{x \to \infty} (4 x + 7) \ln (\frac{x + 1}{x})}$

Step 4

Rewrite $(4 x + 7) \ln (\frac{x + 1}{x})$ as $\frac{\ln (\frac{x + 1}{x})}{\frac{1}{4 x + 7}}$ .

$e^{lim_{x \to \infty} \frac{\ln (\frac{x + 1}{x})}{\frac{1}{4 x + 7}}}$

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.

$e^{\frac{lim_{x \to \infty} \ln (\frac{x + 1}{x})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

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.

$e^{\frac{\ln (lim_{x \to \infty} \frac{x + 1}{x})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.2

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

$e^{\frac{\ln (lim_{x \to \infty} \frac{\frac{x}{x} + \frac{1}{x}}{\frac{x}{x}})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3

Evaluate the limit.

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Step 5.1.2.3.1

Cancel the common factor of $x$ .

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Step 5.1.2.3.1.1

Cancel the common factor.

$e^{\frac{\ln (lim_{x \to \infty} \frac{\frac{x}{x} + \frac{1}{x}}{\frac{x}{x}})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3.1.2

Rewrite the expression.

$e^{\frac{\ln (lim_{x \to \infty} \frac{1 + \frac{1}{x}}{\frac{x}{x}})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3.2

Cancel the common factor of $x$ .

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Step 5.1.2.3.2.1

Cancel the common factor.

$e^{\frac{\ln (lim_{x \to \infty} \frac{1 + \frac{1}{x}}{\frac{x}{x}})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3.2.2

Rewrite the expression.

$e^{\frac{\ln (lim_{x \to \infty} \frac{1 + \frac{1}{x}}{1})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3.3

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

$e^{\frac{\ln (\frac{lim_{x \to \infty} 1 + \frac{1}{x}}{lim_{x \to \infty} 1})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3.4

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

$e^{\frac{\ln (\frac{lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} 1})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.3.5

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

$e^{\frac{\ln (\frac{1 + lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} 1})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.4

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

$e^{\frac{\ln (\frac{1 + 0}{lim_{x \to \infty} 1})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

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 $\infty$ .

$e^{\frac{\ln (\frac{1 + 0}{1})}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.5.2

Simplify the answer.

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Step 5.1.2.5.2.1

Divide $1 + 0$ by $1$ .

$e^{\frac{\ln (1 + 0)}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.5.2.2

Add $1$ and $0$ .

$e^{\frac{\ln (1)}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.2.5.2.3

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

$e^{\frac{0}{lim_{x \to \infty} \frac{1}{4 x + 7}}}$

Step 5.1.3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{4 x + 7}$ 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.

$lim_{x \to \infty} \frac{\ln (\frac{x + 1}{x})}{\frac{1}{4 x + 7}} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (\frac{x + 1}{x})]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$e^{lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (\frac{x + 1}{x})]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

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

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Step 5.3.2.1

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

$e^{lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{x + 1}{x}]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.2.2

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

$e^{lim_{x \to \infty} \frac{\frac{1}{u} \frac{d}{d x} [\frac{x + 1}{x}]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.2.3

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

$e^{lim_{x \to \infty} \frac{\frac{1}{\frac{x + 1}{x}} \frac{d}{d x} [\frac{x + 1}{x}]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.3

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

$e^{lim_{x \to \infty} \frac{1 \frac{x}{x + 1} \frac{d}{d x} [\frac{x + 1}{x}]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.4

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

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \frac{d}{d x} [\frac{x + 1}{x}]}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

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 $f (x) = x + 1$ and $g (x) = x$ .

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x \frac{d}{d x} [x + 1] - (x + 1) \frac{d}{d x} [x]}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.6

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

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x (\frac{d}{d x} [x] + \frac{d}{d x} [1]) - (x + 1) \frac{d}{d x} [x]}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.7

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

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x (1 + \frac{d}{d x} [1]) - (x + 1) \frac{d}{d x} [x]}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.8

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

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x (1 + 0) - (x + 1) \frac{d}{d x} [x]}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.9

Add $1$ and $0$ .

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x \cdot 1 - (x + 1) \frac{d}{d x} [x]}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.10

Multiply $x$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x - (x + 1) \frac{d}{d x} [x]}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.11

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

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x - (x + 1) \cdot 1}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.12

Multiply $- 1$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{x}{x + 1} \cdot \frac{x - (x + 1)}{x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.13

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

$e^{lim_{x \to \infty} \frac{\frac{x (x - (x + 1))}{(x + 1) x^{2}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.14

Cancel the common factors.

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Step 5.3.14.1

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

$e^{lim_{x \to \infty} \frac{\frac{x (x - (x + 1))}{x (x + 1) x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.14.2

Cancel the common factor.

$e^{lim_{x \to \infty} \frac{\frac{x (x - (x + 1))}{x (x + 1) x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.14.3

Rewrite the expression.

$e^{lim_{x \to \infty} \frac{\frac{x - (x + 1)}{(x + 1) x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15

Simplify.

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Step 5.3.15.1

Apply the distributive property.

$e^{lim_{x \to \infty} \frac{\frac{x - x - 1 \cdot 1}{(x + 1) x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.2

Apply the distributive property.

$e^{lim_{x \to \infty} \frac{\frac{x - x - 1 \cdot 1}{x \cdot x + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.3

Simplify the numerator.

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Step 5.3.15.3.1

Subtract $x$ from $x$ .

$e^{lim_{x \to \infty} \frac{\frac{0 - 1 \cdot 1}{x \cdot x + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.3.2

Subtract $1 \cdot 1$ from $0$ .

$e^{lim_{x \to \infty} \frac{\frac{- 1 \cdot 1}{x \cdot x + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.3.3

Multiply $- 1$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{- 1}{x \cdot x + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.4

Combine terms.

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Step 5.3.15.4.1

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

$e^{lim_{x \to \infty} \frac{\frac{- 1}{x^{1} x + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.4.2

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

$e^{lim_{x \to \infty} \frac{\frac{- 1}{x^{1} x^{1} + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.4.3

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

$e^{lim_{x \to \infty} \frac{\frac{- 1}{x^{1 + 1} + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.4.4

Add $1$ and $1$ .

$e^{lim_{x \to \infty} \frac{\frac{- 1}{x^{2} + 1 x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.4.5

Multiply $x$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{- 1}{x^{2} + x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.4.6

Move the negative in front of the fraction.

$e^{lim_{x \to \infty} \frac{- \frac{1}{x^{2} + x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.5

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

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Step 5.3.15.5.1

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x \cdot x + x}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.5.2

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x \cdot x + x^{1}}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.5.3

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x \cdot x + x \cdot 1}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.15.5.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{\frac{d}{d x} [\frac{1}{4 x + 7}]}}$

Step 5.3.16

Rewrite $\frac{1}{4 x + 7}$ as ${(4 x + 7)}^{- 1}$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{\frac{d}{d x} [{(4 x + 7)}^{- 1}]}}$

Step 5.3.17

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = 4 x + 7$ .

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Step 5.3.17.1

To apply the Chain Rule, set $u$ as $4 x + 7$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [4 x + 7]}}$

Step 5.3.17.2

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- u^{- 2} \frac{d}{d x} [4 x + 7]}}$

Step 5.3.17.3

Replace all occurrences of $u$ with $4 x + 7$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} \frac{d}{d x} [4 x + 7]}}$

Step 5.3.18

By the Sum Rule, the derivative of $4 x + 7$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [7]$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} (\frac{d}{d x} [4 x] + \frac{d}{d x} [7])}}$

Step 5.3.19

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} (4 \frac{d}{d x} [x] + \frac{d}{d x} [7])}}$

Step 5.3.20

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} (4 \cdot 1 + \frac{d}{d x} [7])}}$

Step 5.3.21

Multiply $4$ by $1$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} (4 + \frac{d}{d x} [7])}}$

Step 5.3.22

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} (4 + 0)}}$

Step 5.3.23

Add $4$ and $0$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- {(4 x + 7)}^{- 2} \cdot 4}}$

Step 5.3.24

Multiply $4$ by $- 1$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- 4 {(4 x + 7)}^{- 2}}}$

Step 5.3.25

Simplify.

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Step 5.3.25.1

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

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- 4 \frac{1}{{(4 x + 7)}^{2}}}}$

Step 5.3.25.2

Combine terms.

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Step 5.3.25.2.1

Combine $- 4$ and $\frac{1}{{(4 x + 7)}^{2}}$ .

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{\frac{- 4}{{(4 x + 7)}^{2}}}}$

Step 5.3.25.2.2

Move the negative in front of the fraction.

$e^{lim_{x \to \infty} \frac{- \frac{1}{x (x + 1)}}{- \frac{4}{{(4 x + 7)}^{2}}}}$

Step 5.4

Multiply the numerator by the reciprocal of the denominator.

$e^{lim_{x \to \infty} - \frac{1}{x (x + 1)} \cdot - 1 \frac{{(4 x + 7)}^{2}}{4}}$

Step 5.5

Combine factors.

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Step 5.5.1

Multiply $- 1$ by $- 1$ .

$e^{lim_{x \to \infty} 1 \frac{1}{x (x + 1)} \frac{{(4 x + 7)}^{2}}{4}}$

Step 5.5.2

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

$e^{lim_{x \to \infty} \frac{1}{x (x + 1)} \cdot \frac{{(4 x + 7)}^{2}}{4}}$

Step 5.5.3

Multiply $\frac{1}{x (x + 1)}$ by $\frac{{(4 x + 7)}^{2}}{4}$ .

$e^{lim_{x \to \infty} \frac{{(4 x + 7)}^{2}}{x (x + 1) \cdot 4}}$

Step 5.6

Move $4$ to the left of $x (x + 1)$ .

$e^{lim_{x \to \infty} \frac{{(4 x + 7)}^{2}}{4 x (x + 1)}}$

Step 6

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $x$ .

$e^{\frac{1}{4} lim_{x \to \infty} \frac{{(4 x + 7)}^{2}}{x (x + 1)}}$

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

$e^{\frac{1}{4} lim_{x \to \infty} \frac{{(4 x + 7)}^{2}}{x \cdot x + x \cdot 1}}$

Step 7.2

Multiply $x$ by $x$ .

$e^{\frac{1}{4} lim_{x \to \infty} \frac{{(4 x + 7)}^{2}}{x^{2} + x \cdot 1}}$

Step 7.3

Multiply $x$ by $1$ .

$e^{\frac{1}{4} lim_{x \to \infty} \frac{{(4 x + 7)}^{2}}{x^{2} + x}}$

Step 8

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

$e^{\frac{1}{4} lim_{x \to \infty} \frac{{(\frac{4 x}{x} + \frac{7}{x})}^{2}}{\frac{x^{2}}{x^{2}} + \frac{x}{x^{2}}}}$

Step 9

Evaluate the limit.

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Step 9.1

Simplify each term.

$e^{\frac{1}{4} lim_{x \to \infty} \frac{{(\frac{4 x}{x} + \frac{7}{x})}^{2}}{1 + \frac{1}{x}}}$

Step 9.2

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

$e^{\frac{1}{4} \cdot \frac{lim_{x \to \infty} {(\frac{4 x}{x} + \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.3

Move the exponent $2$ from ${(\frac{4 x}{x} + \frac{7}{x})}^{2}$ outside the limit using the Limits Power Rule.

$e^{\frac{1}{4} \cdot \frac{{(lim_{x \to \infty} \frac{4 x}{x} + \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.4

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

$e^{\frac{1}{4} \cdot \frac{{(lim_{x \to \infty} \frac{4 x}{x} + lim_{x \to \infty} \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.5

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$e^{\frac{1}{4} \cdot \frac{{(4 lim_{x \to \infty} \frac{x}{x} + lim_{x \to \infty} \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.6

Cancel the common factor of $x$ .

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Step 9.6.1

Cancel the common factor.

$e^{\frac{1}{4} \cdot \frac{{(4 lim_{x \to \infty} \frac{x}{x} + lim_{x \to \infty} \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.6.2

Rewrite the expression.

$e^{\frac{1}{4} \cdot \frac{{(4 lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.7

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

$e^{\frac{1}{4} \cdot \frac{{(4 \cdot 1 + lim_{x \to \infty} \frac{7}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 9.8

Move the term $7$ outside of the limit because it is constant with respect to $x$ .

$e^{\frac{1}{4} \cdot \frac{{(4 \cdot 1 + 7 lim_{x \to \infty} \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 10

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

$e^{\frac{1}{4} \cdot \frac{{(4 \cdot 1 + 7 \cdot 0)}^{2}}{lim_{x \to \infty} 1 + \frac{1}{x}}}$

Step 11

Evaluate the limit.

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Step 11.1

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

$e^{\frac{1}{4} \cdot \frac{{(4 \cdot 1 + 7 \cdot 0)}^{2}}{lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{1}{x}}}$

Step 11.2

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

$e^{\frac{1}{4} \cdot \frac{{(4 \cdot 1 + 7 \cdot 0)}^{2}}{1 + lim_{x \to \infty} \frac{1}{x}}}$

Step 12

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

$e^{\frac{1}{4} \cdot \frac{{(4 \cdot 1 + 7 \cdot 0)}^{2}}{1 + 0}}$

Step 13

Simplify the answer.

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Step 13.1

Simplify the numerator.

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Step 13.1.1

Multiply $4$ by $1$ .

$e^{\frac{1}{4} \cdot \frac{{(4 + 7 \cdot 0)}^{2}}{1 + 0}}$

Step 13.1.2

Multiply $7$ by $0$ .

$e^{\frac{1}{4} \cdot \frac{{(4 + 0)}^{2}}{1 + 0}}$

Step 13.1.3

Add $4$ and $0$ .

$e^{\frac{1}{4} \cdot \frac{4^{2}}{1 + 0}}$

Step 13.1.4

Raise $4$ to the power of $2$ .

$e^{\frac{1}{4} \cdot \frac{16}{1 + 0}}$

Step 13.2

Add $1$ and $0$ .

$e^{\frac{1}{4} \cdot \frac{16}{1}}$

Step 13.3

Cancel the common factor of $4$ .

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Step 13.3.1

Factor $4$ out of $16$ .

$e^{\frac{1}{4} \cdot \frac{4 (4)}{1}}$

Step 13.3.2

Cancel the common factor.

$e^{\frac{1}{4} \cdot \frac{4 \cdot 4}{1}}$

Step 13.3.3

Rewrite the expression.

$e^{4}$

Step 14

The result can be shown in multiple forms.

Exact Form:

$e^{4}$

Decimal Form:

$54.59815003 \dots$