Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

Move the limit inside the logarithm.

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

Step 1.2.1.2

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

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

Step 1.2.1.3

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

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

Step 1.2.1.4

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

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

Step 1.2.2

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

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

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Multiply $a$ by $0$ .

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

Step 1.2.3.2

Add $1$ and $0$ .

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

Step 1.2.3.3

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

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

Step 1.3

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

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

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

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

Step 3.4

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

$lim_{x \to \infty} \frac{\frac{1}{1 + \frac{a}{x}} (0 + \frac{d}{d x} [\frac{a}{x}])}{\frac{d}{d x} [\frac{1}{x}]}$

Step 3.5

Add $0$ and $\frac{d}{d x} [\frac{a}{x}]$ .

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

Step 3.6

Since $a$ is constant with respect to $x$ , the derivative of $\frac{a}{x}$ with respect to $x$ is $a \frac{d}{d x} [\frac{1}{x}]$ .

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

Step 3.7

Combine $a$ and $\frac{1}{1 + \frac{a}{x}}$ .

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

Step 3.8

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

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

Step 3.9

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

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

Step 3.10

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

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

Step 3.11

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.12

Simplify.

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Step 3.12.1

Apply the distributive property.

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

Step 3.12.2

Combine terms.

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Step 3.12.2.1

Multiply $x^{2}$ by $1$ .

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

Step 3.12.2.2

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

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

Step 3.12.2.3

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

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Step 3.12.2.3.1

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

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

Step 3.12.2.3.2

Cancel the common factors.

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Step 3.12.2.3.2.1

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

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

Step 3.12.2.3.2.2

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

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

Step 3.12.2.3.2.3

Cancel the common factor.

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

Step 3.12.2.3.2.4

Rewrite the expression.

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

Step 3.12.2.3.2.5

Divide $a x$ by $1$ .

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

Step 3.12.3

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

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Step 3.12.3.1

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

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

Step 3.12.3.2

Factor $x$ out of $a x$ .

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

Step 3.12.3.3

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

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

Step 3.13

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

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

Step 3.14

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

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

Step 3.15

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

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

Combine factors.

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Step 5.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2

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

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

Step 5.3

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

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

Step 6

Evaluate the limit.

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Step 6.1

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

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Step 6.1.1

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

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

Step 6.1.2

Cancel the common factors.

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Step 6.1.2.1

Cancel the common factor.

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

Step 6.1.2.2

Rewrite the expression.

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

Step 6.2

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

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

Step 7

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

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

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.

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

Step 8.1.2

Rewrite the expression.

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

Step 8.2

Cancel the common factor of $x$ .

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Step 8.2.1

Cancel the common factor.

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

Step 8.2.2

Rewrite the expression.

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

Step 8.3

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

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

Step 8.4

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

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

Step 8.5

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

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

Step 8.6

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

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

Step 8.7

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

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

Step 9

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

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

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 $0$ .

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

Step 10.1.2

Add $1$ and $0$ .

$a \frac{1}{1}$

Step 10.2

Divide $1$ by $1$ .

$a \cdot 1$

Step 10.3

Multiply $a$ by $1$ .

$a$