Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

Move the limit into the exponent.

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

Step 3

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

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

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.

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

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.

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Evaluate the limit.

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Step 4.1.2.3.1

Cancel the common factor of $x$ .

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Step 4.1.2.3.1.1

Cancel the common factor.

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

Step 4.1.2.3.1.2

Rewrite the expression.

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

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.

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

Step 4.1.2.3.2.2

Rewrite the expression.

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

Step 4.1.2.3.3

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

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

Step 4.1.2.3.4

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

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

Step 4.1.2.3.5

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

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

Step 4.1.2.4

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

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

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 $\infty$ .

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

Step 4.1.2.5.2

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

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

Step 4.1.2.6

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

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

Step 4.1.2.7

Simplify the answer.

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Step 4.1.2.7.1

Add $1$ and $0$ .

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

Step 4.1.2.7.2

Simplify the denominator.

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Step 4.1.2.7.2.1

Multiply $- 1$ by $0$ .

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

Step 4.1.2.7.2.2

Add $1$ and $0$ .

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

Step 4.1.2.7.3

Divide $1$ by $1$ .

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

Step 4.1.2.7.4

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

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

Step 4.1.3

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

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

Step 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.

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

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

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

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

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Step 4.3.2.1

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

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

Step 4.3.2.2

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

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

Step 4.3.2.3

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

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

Step 4.3.3

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

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

Step 4.3.4

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

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

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

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

Step 4.3.6

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

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

Step 4.3.7

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

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

Step 4.3.8

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

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

Step 4.3.9

Add $1$ and $0$ .

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

Step 4.3.10

Multiply $x - 1$ by $1$ .

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

Step 4.3.11

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

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

Step 4.3.12

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

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

Step 4.3.13

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

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

Step 4.3.14

Add $1$ and $0$ .

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

Step 4.3.15

Multiply $- 1$ by $1$ .

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

Step 4.3.16

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

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

Step 4.3.17

Cancel the common factors.

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Step 4.3.17.1

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

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

Step 4.3.17.2

Cancel the common factor.

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

Step 4.3.17.3

Rewrite the expression.

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

Step 4.3.18

Simplify.

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Step 4.3.18.1

Apply the distributive property.

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

Step 4.3.18.2

Simplify the numerator.

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Step 4.3.18.2.1

Combine the opposite terms in $x - 1 - x - 1 \cdot 1$ .

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Step 4.3.18.2.1.1

Subtract $x$ from $x$ .

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

Step 4.3.18.2.1.2

Subtract $1$ from $0$ .

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

Step 4.3.18.2.2

Multiply $- 1$ by $1$ .

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

Step 4.3.18.2.3

Subtract $1$ from $- 1$ .

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

Step 4.3.18.3

Move the negative in front of the fraction.

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

Step 4.3.19

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

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

Step 4.3.20

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

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

Step 4.3.21

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

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

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.5

Combine factors.

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Step 4.5.1

Multiply $- 1$ by $- 1$ .

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

Step 4.5.2

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

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

Step 4.5.3

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

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

Step 5

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

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

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.

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

Step 6.1.2

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

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

Step 6.1.3

Evaluate the limit of the denominator.

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Step 6.1.3.1

Apply the distributive property.

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

Step 6.1.3.2

Apply the distributive property.

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

Step 6.1.3.3

Apply the distributive property.

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

Step 6.1.3.4

Reorder $x$ and $- 1$ .

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

Step 6.1.3.5

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

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

Step 6.1.3.6

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

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

Step 6.1.3.7

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

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

Step 6.1.3.8

Simplify by adding terms.

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Step 6.1.3.8.1

Add $1$ and $1$ .

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

Step 6.1.3.8.2

Simplify.

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Step 6.1.3.8.2.1

Multiply $x$ by $1$ .

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

Step 6.1.3.8.2.2

Multiply $- 1$ by $1$ .

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

Step 6.1.3.8.3

Add $- 1 x$ and $x$ .

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

Step 6.1.3.8.4

Subtract $1$ from $0$ .

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

Step 6.1.3.9

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

$e^{2 \frac{\infty}{\infty}}$

Step 6.1.3.10

Infinity divided by infinity is undefined.

Undefined

$e^{2 \frac{\infty}{\infty}}$

Step 6.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{2 \frac{\infty}{\infty}}$

Step 6.2

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.

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

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

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

Step 6.3.2

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

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

Step 6.3.3

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

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

Step 6.3.4

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

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

Step 6.3.5

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

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

Step 6.3.6

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

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

Step 6.3.7

Add $1$ and $0$ .

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

Step 6.3.8

Multiply $x + 1$ by $1$ .

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

Step 6.3.9

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

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

Step 6.3.10

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

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

Step 6.3.11

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

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

Step 6.3.12

Add $1$ and $0$ .

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

Step 6.3.13

Multiply $x - 1$ by $1$ .

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

Step 6.3.14

Add $x$ and $x$ .

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

Step 6.3.15

Subtract $1$ from $1$ .

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

Step 6.3.16

Add $2 x$ and $0$ .

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

Step 6.4

Reduce.

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Step 6.4.1

Cancel the common factor of $2$ .

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Step 6.4.1.1

Cancel the common factor.

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

Step 6.4.1.2

Rewrite the expression.

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

Step 6.4.2

Cancel the common factor of $x$ .

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Step 6.4.2.1

Cancel the common factor.

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

Step 6.4.2.2

Rewrite the expression.

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

Step 7

Evaluate the limit.

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Step 7.1

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

$e^{2 \cdot 1}$

Step 7.2

Multiply $2$ by $1$ .

$e^{2}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$e^{2}$

Decimal Form:

$7.38905609 \dots$