Calculus Examples

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

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

Move the limit inside the logarithm.

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

Step 1.1.2.1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

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

Step 1.1.2.1.3

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

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

Step 1.1.2.1.4

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

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

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Raising $0$ to any positive power yields $0$ .

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

Step 1.1.2.3.2

Add $1$ and $0$ .

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

Step 1.1.2.3.3

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

$\frac{0}{lim_{x \to 0^{+}} x^{3}}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Move the exponent $3$ from $x^{3}$ outside the limit using the Limits Power Rule.

$\frac{0}{{(lim_{x \to 0^{+}} x)}^{3}}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{0}{0^{3}}$

Step 1.1.3.3

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 0^{+}} \frac{\ln (1 + x^{2})}{x^{3}} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} [\ln (1 + x^{2})]}{\frac{d}{d x} [x^{3}]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0^{+}} \frac{\frac{d}{d x} [\ln (1 + x^{2})]}{\frac{d}{d x} [x^{3}]}$

Step 1.3.2

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 $g (x) = 1 + x^{2}$ .

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Step 1.3.2.1

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

$lim_{x \to 0^{+}} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [1 + x^{2}]}{\frac{d}{d x} [x^{3}]}$

Step 1.3.2.2

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

$lim_{x \to 0^{+}} \frac{\frac{1}{u} \frac{d}{d x} [1 + x^{2}]}{\frac{d}{d x} [x^{3}]}$

Step 1.3.2.3

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

$lim_{x \to 0^{+}} \frac{\frac{1}{1 + x^{2}} \frac{d}{d x} [1 + x^{2}]}{\frac{d}{d x} [x^{3}]}$

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

$lim_{x \to 0^{+}} \frac{\frac{1}{1 + x^{2}} \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [x^{3}]}$

Step 1.3.6

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

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

Step 1.3.7

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

$lim_{x \to 0^{+}} \frac{\frac{2}{1 + x^{2}} x}{\frac{d}{d x} [x^{3}]}$

Step 1.3.8

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

$lim_{x \to 0^{+}} \frac{\frac{2 x}{1 + x^{2}}}{\frac{d}{d x} [x^{3}]}$

Step 1.3.9

Reorder terms.

$lim_{x \to 0^{+}} \frac{\frac{2 x}{x^{2} + 1}}{\frac{d}{d x} [x^{3}]}$

Step 1.3.10

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$lim_{x \to 0^{+}} \frac{\frac{2 x}{x^{2} + 1}}{3 x^{2}}$

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 0^{+}} \frac{2 x}{x^{2} + 1} \cdot \frac{1}{3 x^{2}}$

Step 1.5

Multiply $\frac{2 x}{x^{2} + 1}$ by $\frac{1}{3 x^{2}}$ .

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

Step 1.6

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

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Step 1.6.1

Factor $x$ out of $2 x$ .

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

Step 1.6.2

Cancel the common factors.

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Step 1.6.2.1

Factor $x$ out of $(x^{2} + 1) (3 x^{2})$ .

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

Step 1.6.2.2

Cancel the common factor.

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

Step 1.6.2.3

Rewrite the expression.

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

Step 2

Since the numerator is positive and the denominator $(x^{2} + 1) (3 x)$ approaches zero and is greater than zero for $x$ near $0$ to the right, the function increases without bound.

$\infty$