Evaluate limit as x approaches infinity of (1+a/x)^(bx)

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

Combine terms.

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

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

Combine the numerators over the common denominator.

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

Use the properties of logarithms to simplify the limit.

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

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

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

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

Move the limit into the exponent.

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

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

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

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

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

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

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

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

Evaluate the limit of the numerator.

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Take the limit of each term.

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Move the limit inside the logarithm.

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

Reorder $x$ and $a$ .

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

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

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

Cancel the common factor of $x$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

Cancel the common factor of $x$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

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

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

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

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

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

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

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

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

Evaluate the limits by plugging in the value for the variable.

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

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

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

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

Simplify the answer.

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

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

Multiply $a$ by $0$ .

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

Add $0$ and $1$ .

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

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

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

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

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

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

Undefined

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

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.

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

Find the derivative of the numerator and denominator.

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Differentiate the numerator and denominator.

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

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

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

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

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

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

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

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

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

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

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

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 $g (x) = x$ .

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

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

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

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

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

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

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

Add $1$ and $0$ .

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

Multiply $x$ by $1$ .

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

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

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

Multiply $- 1$ by $1$ .

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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Apply the distributive property.

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

Apply the distributive property.

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

Simplify the numerator.

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Subtract $x$ from $x$ .

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

Subtract $a$ from $0$ .

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

Combine terms.

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

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

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

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

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

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

Add $1$ and $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Combine factors.

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Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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

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

Factor $x$ out of $a x$ .

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

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

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

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

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

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

Cancel the common factors.

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Cancel the common factor.

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

Rewrite the expression.

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

Take the limit of each term.

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

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

Reorder $x$ and $a$ .

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

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

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

Cancel the common factor of $x$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

Cancel the common factor of $x$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

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

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

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

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

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

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

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

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

Evaluate the limits by plugging in the value for the variable.

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

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

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

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

Simplify the answer.

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Simplify the denominator.

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Multiply $a$ by $0$ .

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

Add $0$ and $1$ .

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

Divide $1$ by $1$ .

$e^{b a \cdot 1}$

Multiply $a$ by $1$ .

$e^{b a}$

Evaluate limit as x approaches infinity of (1+a/x)^(bx)

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