Calculus Examples

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

Step 1

Combine terms.

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Step 1.1

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

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

Step 1.2

Combine the numerators over the common denominator.

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

Step 2

Use the properties of logarithms to simplify the limit.

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Step 2.1

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

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

Step 2.2

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

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

Step 3

Move the limit into the exponent.

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

Step 4

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

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

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.

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

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.

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Evaluate the limit.

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Step 5.1.2.3.1

Cancel the common factor of $x$ .

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Step 5.1.2.3.1.1

Cancel the common factor.

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

Step 5.1.2.3.1.2

Rewrite the expression.

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

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.

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

Step 5.1.2.3.2.2

Rewrite the expression.

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

Step 5.1.2.3.3

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

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

Step 5.1.2.3.4

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

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

Step 5.1.2.3.5

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

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

Step 5.1.2.3.6

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

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

Step 5.1.2.4

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

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

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 $\infty$ .

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

Step 5.1.2.5.2

Simplify the answer.

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Step 5.1.2.5.2.1

Divide $1 + a \cdot 0$ by $1$ .

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

Step 5.1.2.5.2.2

Multiply $a$ by $0$ .

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

Step 5.1.2.5.2.3

Add $1$ and $0$ .

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

Step 5.1.2.5.2.4

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.

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

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

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Step 5.3.2.1

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

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

Step 5.3.2.2

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

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

Step 5.3.2.3

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

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

Step 5.3.3

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

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

Step 5.3.4

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

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

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

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

Step 5.3.6

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

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

Step 5.3.7

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

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

Step 5.3.8

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

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

Step 5.3.9

Add $1$ and $0$ .

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

Step 5.3.10

Multiply $x$ by $1$ .

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

Step 5.3.11

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

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

Step 5.3.12

Multiply $- 1$ by $1$ .

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

Step 5.3.13

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

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

Step 5.3.14

Cancel the common factors.

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Step 5.3.14.1

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

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

Step 5.3.14.2

Cancel the common factor.

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

Step 5.3.14.3

Rewrite the expression.

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

Step 5.3.15

Simplify.

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Step 5.3.15.1

Apply the distributive property.

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

Step 5.3.15.2

Apply the distributive property.

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

Step 5.3.15.3

Simplify the numerator.

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Step 5.3.15.3.1

Subtract $x$ from $x$ .

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

Step 5.3.15.3.2

Subtract $a$ from $0$ .

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

Step 5.3.15.4

Combine terms.

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Step 5.3.15.4.1

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

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

Step 5.3.15.4.2

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

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

Step 5.3.15.4.3

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

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

Step 5.3.15.4.4

Add $1$ and $1$ .

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

Step 5.3.15.4.5

Move the negative in front of the fraction.

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

Step 5.3.15.5

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

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Step 5.3.15.5.1

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

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

Step 5.3.15.5.2

Factor $x$ out of $a x$ .

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

Step 5.3.15.5.3

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

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

Step 5.3.16

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

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

Step 5.3.17

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

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

Step 5.3.18

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

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

Step 5.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.5

Combine factors.

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Step 5.5.1

Multiply $- 1$ by $- 1$ .

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

Step 5.5.2

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

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

Step 5.5.3

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

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

Step 5.6

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

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Step 5.6.1

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

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

Step 5.6.2

Cancel the common factors.

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Step 5.6.2.1

Cancel the common factor.

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

Step 5.6.2.2

Rewrite the expression.

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

Step 6

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

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

Step 7

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

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

Step 8

Evaluate the limit.

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Step 8.1

Cancel the common factor of $x$ .

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Step 8.1.1

Cancel the common factor.

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

Step 8.1.2

Rewrite the expression.

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

Step 8.2

Cancel the common factor of $x$ .

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Step 8.2.1

Cancel the common factor.

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

Step 8.2.2

Rewrite the expression.

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

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

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

Step 8.7

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

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

Step 9

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

$e^{a \frac{1}{1 + a \cdot 0}}$

Step 10

Simplify the answer.

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Step 10.1

Simplify the denominator.

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Step 10.1.1

Multiply $a$ by $0$ .

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

Step 10.1.2

Add $1$ and $0$ .

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

Step 10.2

Divide $1$ by $1$ .

$e^{a \cdot 1}$

Step 10.3

Multiply $a$ by $1$ .

$e^{a}$