Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Move the limit under the radical sign.

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Rewrite $0$ as $0^{2}$ .

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

Step 1.1.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

Move the limit inside the logarithm.

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

Step 1.1.3.1.2

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

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

$\frac{0}{\ln (0 + 1)}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Add $0$ and $1$ .

$\frac{0}{\ln (1)}$

Step 1.1.3.3.2

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

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Step 1.3.3

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

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

Step 1.3.4

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.3.5

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 1.3.6

Combine the numerators over the common denominator.

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

Step 1.3.7

Simplify the numerator.

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Step 1.3.7.1

Multiply $- 1$ by $2$ .

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

Step 1.3.7.2

Subtract $2$ from $1$ .

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

Step 1.3.8

Move the negative in front of the fraction.

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

Step 1.3.9

Simplify.

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Step 1.3.9.1

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

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

Step 1.3.9.2

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

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

Step 1.3.10

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 $g (x) = x + 1$ .

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Step 1.3.10.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 1.3.10.2

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

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

Step 1.3.10.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 1.3.11

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

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

Step 1.3.12

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

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

Step 1.3.13

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

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

Step 1.3.14

Add $1$ and $0$ .

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

Step 1.3.15

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

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

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5

Rewrite $x^{\frac{1}{2}}$ as $\sqrt{x}$ .

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

Step 1.6

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

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

Step 2

Since the numerator is positive and the denominator $2 \sqrt{x}$ approaches zero and is greater than zero for $x$ near $0$ to the right, the function increases without bound.

$\infty$