Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(\frac{x - 4}{x + 3})}^{2 x + 1}$ as $e^{\ln ({(\frac{x - 4}{x + 3})}^{2 x + 1})}$ .

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

Step 1.2

Expand $\ln ({(\frac{x - 4}{x + 3})}^{2 x + 1})$ by moving $2 x + 1$ outside the logarithm.

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

Step 2

Move the limit into the exponent.

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

Step 3

Rewrite $(2 x + 1) \ln (\frac{x - 4}{x + 3})$ as $\frac{\ln (\frac{x - 4}{x + 3})}{\frac{1}{2 x + 1}}$ .

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

Step 4

Apply L'Hospital's rule.

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Step 4.1

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

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Step 4.1.1

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

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

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

Move the limit inside the logarithm.

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Evaluate the limit.

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Step 4.1.2.3.1

Simplify each term.

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Step 4.1.2.3.1.1

Cancel the common factor of $x$ .

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Step 4.1.2.3.1.1.1

Cancel the common factor.

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

Step 4.1.2.3.1.1.2

Rewrite the expression.

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

Step 4.1.2.3.1.2

Move the negative in front of the fraction.

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

Step 4.1.2.3.2

Cancel the common factor of $x$ .

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Step 4.1.2.3.2.1

Cancel the common factor.

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

Step 4.1.2.3.2.2

Rewrite the expression.

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

Step 4.1.2.3.3

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

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

Step 4.1.2.3.4

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

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

Step 4.1.2.3.5

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

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

Step 4.1.2.3.6

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

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

Step 4.1.2.4

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

$e^{\frac{\ln (\frac{1 - 4 \cdot 0}{lim_{x \to \infty} 1 + \frac{3}{x}})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.5

Evaluate the limit.

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Step 4.1.2.5.1

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

$e^{\frac{\ln (\frac{1 - 4 \cdot 0}{lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{3}{x}})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.5.2

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

$e^{\frac{\ln (\frac{1 - 4 \cdot 0}{1 + lim_{x \to \infty} \frac{3}{x}})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.5.3

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$e^{\frac{\ln (\frac{1 - 4 \cdot 0}{1 + 3 lim_{x \to \infty} \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.6

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

$e^{\frac{\ln (\frac{1 - 4 \cdot 0}{1 + 3 \cdot 0})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.7

Simplify the answer.

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Step 4.1.2.7.1

Simplify the numerator.

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Step 4.1.2.7.1.1

Multiply $- 4$ by $0$ .

$e^{\frac{\ln (\frac{1 + 0}{1 + 3 \cdot 0})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.7.1.2

Add $1$ and $0$ .

$e^{\frac{\ln (\frac{1}{1 + 3 \cdot 0})}{lim_{x \to \infty} \frac{1}{2 x + 1}}}$

Step 4.1.2.7.2

Simplify the denominator.

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Step 4.1.2.7.2.1

Multiply $3$ by $0$ .

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

Step 4.1.2.7.2.2

Add $1$ and $0$ .

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

Step 4.1.2.7.3

Divide $1$ by $1$ .

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

Step 4.1.2.7.4

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

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

Step 4.1.3

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

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

Step 4.1.4

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

Undefined

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

Step 4.2

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.

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

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

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

Step 4.3.2

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 $g (x) = \frac{x - 4}{x + 3}$ .

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Step 4.3.2.1

To apply the Chain Rule, set $u$ as $\frac{x - 4}{x + 3}$ .

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

Step 4.3.2.2

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

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

Step 4.3.2.3

Replace all occurrences of $u$ with $\frac{x - 4}{x + 3}$ .

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

Step 4.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{x - 4}{x + 3}$ .

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

Step 4.3.4

Multiply $\frac{x + 3}{x - 4}$ by $1$ .

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

Step 4.3.5

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 $f (x) = x - 4$ and $g (x) = x + 3$ .

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

Step 4.3.6

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

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

Step 4.3.7

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

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

Step 4.3.8

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$e^{lim_{x \to \infty} \frac{\frac{x + 3}{x - 4} \cdot \frac{(x + 3) (1 + 0) - (x - 4) \frac{d}{d x} [x + 3]}{{(x + 3)}^{2}}}{\frac{d}{d x} [\frac{1}{2 x + 1}]}}$

Step 4.3.9

Add $1$ and $0$ .

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

Step 4.3.10

Multiply $x + 3$ by $1$ .

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

Step 4.3.11

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

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

Step 4.3.12

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

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

Step 4.3.13

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$e^{lim_{x \to \infty} \frac{\frac{x + 3}{x - 4} \cdot \frac{x + 3 - (x - 4) (1 + 0)}{{(x + 3)}^{2}}}{\frac{d}{d x} [\frac{1}{2 x + 1}]}}$

Step 4.3.14

Add $1$ and $0$ .

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

Step 4.3.15

Multiply $- 1$ by $1$ .

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

Step 4.3.16

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

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

Step 4.3.17

Cancel the common factors.

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Step 4.3.17.1

Factor $x + 3$ out of $(x - 4) {(x + 3)}^{2}$ .

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

Step 4.3.17.2

Cancel the common factor.

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

Step 4.3.17.3

Rewrite the expression.

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

Step 4.3.18

Simplify.

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Step 4.3.18.1

Apply the distributive property.

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

Step 4.3.18.2

Simplify the numerator.

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Step 4.3.18.2.1

Combine the opposite terms in $x + 3 - x - - 4$ .

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Step 4.3.18.2.1.1

Subtract $x$ from $x$ .

$e^{lim_{x \to \infty} \frac{\frac{0 + 3 - - 4}{(x - 4) (x + 3)}}{\frac{d}{d x} [\frac{1}{2 x + 1}]}}$

Step 4.3.18.2.1.2

Add $0$ and $3$ .

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

Step 4.3.18.2.2

Multiply $- 1$ by $- 4$ .

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

Step 4.3.18.2.3

Add $3$ and $4$ .

$e^{lim_{x \to \infty} \frac{\frac{7}{(x - 4) (x + 3)}}{\frac{d}{d x} [\frac{1}{2 x + 1}]}}$

Step 4.3.18.3

Reorder terms.

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{\frac{d}{d x} [\frac{1}{2 x + 1}]}}$

Step 4.3.19

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{\frac{d}{d x} [{(2 x + 1)}^{- 1}]}}$

Step 4.3.20

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 $g (x) = 2 x + 1$ .

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Step 4.3.20.1

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [2 x + 1]}}$

Step 4.3.20.2

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- u^{- 2} \frac{d}{d x} [2 x + 1]}}$

Step 4.3.20.3

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} \frac{d}{d x} [2 x + 1]}}$

Step 4.3.21

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} (\frac{d}{d x} [2 x] + \frac{d}{d x} [1])}}$

Step 4.3.22

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} (2 \frac{d}{d x} [x] + \frac{d}{d x} [1])}}$

Step 4.3.23

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} (2 \cdot 1 + \frac{d}{d x} [1])}}$

Step 4.3.24

Multiply $2$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} (2 + \frac{d}{d x} [1])}}$

Step 4.3.25

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} (2 + 0)}}$

Step 4.3.26

Add $2$ and $0$ .

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- {(2 x + 1)}^{- 2} \cdot 2}}$

Step 4.3.27

Multiply $2$ by $- 1$ .

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- 2 {(2 x + 1)}^{- 2}}}$

Step 4.3.28

Simplify.

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Step 4.3.28.1

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- 2 \frac{1}{{(2 x + 1)}^{2}}}}$

Step 4.3.28.2

Combine terms.

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Step 4.3.28.2.1

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

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{\frac{- 2}{{(2 x + 1)}^{2}}}}$

Step 4.3.28.2.2

Move the negative in front of the fraction.

$e^{lim_{x \to \infty} \frac{\frac{7}{(x + 3) (x - 4)}}{- \frac{2}{{(2 x + 1)}^{2}}}}$

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

$e^{lim_{x \to \infty} \frac{7}{(x + 3) (x - 4)} \cdot - 1 \frac{{(2 x + 1)}^{2}}{2}}$

Step 4.5

Multiply $\frac{7}{(x + 3) (x - 4)}$ by $\frac{{(2 x + 1)}^{2}}{2}$ .

$e^{lim_{x \to \infty} - \frac{7 {(2 x + 1)}^{2}}{(x + 3) (x - 4) \cdot 2}}$

Step 4.6

Move $2$ to the left of $(x + 3) (x - 4)$ .

$e^{lim_{x \to \infty} - \frac{7 {(2 x + 1)}^{2}}{2 (x + 3) (x - 4)}}$

Step 5

Evaluate the limit.

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Step 5.1

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

$e^{- lim_{x \to \infty} \frac{7 {(2 x + 1)}^{2}}{2 (x + 3) (x - 4)}}$

Step 5.2

Move the term $\frac{7}{2}$ outside of the limit because it is constant with respect to $x$ .

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{(x + 3) (x - 4)}}$

Step 6

Simplify.

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Step 6.1

Expand $(x + 3) (x - 4)$ using the FOIL Method.

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Step 6.1.1

Apply the distributive property.

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x (x - 4) + 3 (x - 4)}}$

Step 6.1.2

Apply the distributive property.

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x \cdot x + x \cdot - 4 + 3 (x - 4)}}$

Step 6.1.3

Apply the distributive property.

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x \cdot x + x \cdot - 4 + 3 x + 3 \cdot - 4}}$

Step 6.2

Simplify and combine like terms.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $x$ by $x$ .

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x^{2} + x \cdot - 4 + 3 x + 3 \cdot - 4}}$

Step 6.2.1.2

Move $- 4$ to the left of $x$ .

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x^{2} - 4 \cdot x + 3 x + 3 \cdot - 4}}$

Step 6.2.1.3

Multiply $3$ by $- 4$ .

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x^{2} - 4 x + 3 x - 12}}$

Step 6.2.2

Add $- 4 x$ and $3 x$ .

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(2 x + 1)}^{2}}{x^{2} - x - 12}}$

Step 7

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

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(\frac{2 x}{x} + \frac{1}{x})}^{2}}{\frac{x^{2}}{x^{2}} - \frac{x}{x^{2}} + \frac{- 12}{x^{2}}}}$

Step 8

Evaluate the limit.

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Step 8.1

Simplify each term.

$e^{- \frac{7}{2} lim_{x \to \infty} \frac{{(\frac{2 x}{x} + \frac{1}{x})}^{2}}{1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.2

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

$e^{- \frac{7}{2} \cdot \frac{lim_{x \to \infty} {(\frac{2 x}{x} + \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.3

Move the exponent $2$ from ${(\frac{2 x}{x} + \frac{1}{x})}^{2}$ outside the limit using the Limits Power Rule.

$e^{- \frac{7}{2} \cdot \frac{{(lim_{x \to \infty} \frac{2 x}{x} + \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.4

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

$e^{- \frac{7}{2} \cdot \frac{{(lim_{x \to \infty} \frac{2 x}{x} + lim_{x \to \infty} \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.5

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

$e^{- \frac{7}{2} \cdot \frac{{(2 lim_{x \to \infty} \frac{x}{x} + lim_{x \to \infty} \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.6

Cancel the common factor of $x$ .

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Step 8.6.1

Cancel the common factor.

$e^{- \frac{7}{2} \cdot \frac{{(2 lim_{x \to \infty} \frac{x}{x} + lim_{x \to \infty} \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.6.2

Rewrite the expression.

$e^{- \frac{7}{2} \cdot \frac{{(2 lim_{x \to \infty} 1 + lim_{x \to \infty} \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 8.7

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

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + lim_{x \to \infty} \frac{1}{x})}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

Step 9

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

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + 0)}^{2}}{lim_{x \to \infty} 1 - \frac{1}{x} - \frac{12}{x^{2}}}}$

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 $\infty$ .

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + 0)}^{2}}{lim_{x \to \infty} 1 - lim_{x \to \infty} \frac{1}{x} - lim_{x \to \infty} \frac{12}{x^{2}}}}$

Step 10.2

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

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + 0)}^{2}}{1 - lim_{x \to \infty} \frac{1}{x} - lim_{x \to \infty} \frac{12}{x^{2}}}}$

Step 11

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

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + 0)}^{2}}{1 - 0 - lim_{x \to \infty} \frac{12}{x^{2}}}}$

Step 12

Move the term $12$ outside of the limit because it is constant with respect to $x$ .

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + 0)}^{2}}{1 - 0 - 12 lim_{x \to \infty} \frac{1}{x^{2}}}}$

Step 13

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

$e^{- \frac{7}{2} \cdot \frac{{(2 \cdot 1 + 0)}^{2}}{1 - 0 - 12 \cdot 0}}$

Step 14

Simplify the answer.

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Step 14.1

Simplify the numerator.

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Step 14.1.1

Multiply $2$ by $1$ .

$e^{- \frac{7}{2} \cdot \frac{{(2 + 0)}^{2}}{1 - 0 - 12 \cdot 0}}$

Step 14.1.2

Add $2$ and $0$ .

$e^{- \frac{7}{2} \cdot \frac{2^{2}}{1 - 0 - 12 \cdot 0}}$

Step 14.1.3

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

$e^{- \frac{7}{2} \cdot \frac{4}{1 - 0 - 12 \cdot 0}}$

Step 14.2

Simplify the denominator.

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Step 14.2.1

Multiply $- 1$ by $0$ .

$e^{- \frac{7}{2} \cdot \frac{4}{1 + 0 - 12 \cdot 0}}$

Step 14.2.2

Multiply $- 12$ by $0$ .

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

Step 14.2.3

Add $1 + 0$ and $0$ .

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

Step 14.2.4

Add $1$ and $0$ .

$e^{- \frac{7}{2} \cdot \frac{4}{1}}$

Step 14.3

Cancel the common factor of $2$ .

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Step 14.3.1

Move the leading negative in $- \frac{7}{2}$ into the numerator.

$e^{\frac{- 7}{2} \cdot \frac{4}{1}}$

Step 14.3.2

Factor $2$ out of $4$ .

$e^{\frac{- 7}{2} \cdot \frac{2 (2)}{1}}$

Step 14.3.3

Cancel the common factor.

$e^{\frac{- 7}{2} \cdot \frac{2 \cdot 2}{1}}$

Step 14.3.4

Rewrite the expression.

$e^{- 7 \cdot 2}$

Step 14.4

Multiply $- 7$ by $2$ .

$e^{- 14}$

Step 15

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

$\frac{1}{e^{14}}$