Calculus Examples

$lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{x - 2}$

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.

$\frac{lim_{x \to 2} \sqrt{2 + x} - 2}{lim_{x \to 2} x - 2}$

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

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{lim_{x \to 2} \sqrt{2 + x} - lim_{x \to 2} 2}{lim_{x \to 2} x - 2}$

Step 1.2.1.2

Move the limit under the radical sign.

$\frac{\sqrt{lim_{x \to 2} 2 + x} - lim_{x \to 2} 2}{lim_{x \to 2} x - 2}$

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{\sqrt{lim_{x \to 2} 2 + lim_{x \to 2} x} - lim_{x \to 2} 2}{lim_{x \to 2} x - 2}$

Step 1.2.1.4

Evaluate the limit of $2$ which is constant as $x$ approaches $2$ .

$\frac{\sqrt{2 + lim_{x \to 2} x} - lim_{x \to 2} 2}{lim_{x \to 2} x - 2}$

Step 1.2.1.5

Evaluate the limit of $2$ which is constant as $x$ approaches $2$ .

$\frac{\sqrt{2 + lim_{x \to 2} x} - 1 \cdot 2}{lim_{x \to 2} x - 2}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{\sqrt{2 + 2} - 1 \cdot 2}{lim_{x \to 2} x - 2}$

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Add $2$ and $2$ .

$\frac{\sqrt{4} - 1 \cdot 2}{lim_{x \to 2} x - 2}$

Step 1.2.3.1.2

Rewrite $4$ as $2^{2}$ .

$\frac{\sqrt{2^{2}} - 1 \cdot 2}{lim_{x \to 2} x - 2}$

Step 1.2.3.1.3

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

$\frac{2 - 1 \cdot 2}{lim_{x \to 2} x - 2}$

Step 1.2.3.1.4

Multiply $- 1$ by $2$ .

$\frac{2 - 2}{lim_{x \to 2} x - 2}$

Step 1.2.3.2

Subtract $2$ from $2$ .

$\frac{0}{lim_{x \to 2} x - 2}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{lim_{x \to 2} x - lim_{x \to 2} 2}$

Step 1.3.1.2

Evaluate the limit of $2$ which is constant as $x$ approaches $2$ .

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

Step 1.3.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{2 - 1 \cdot 2}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Multiply $- 1$ by $2$ .

$\frac{0}{2 - 2}$

Step 1.3.3.2

Subtract $2$ from $2$ .

$\frac{0}{0}$

Step 1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 2} \frac{\sqrt{2 + x} - 2}{x - 2} = lim_{x \to 2} \frac{\frac{d}{d x} [\sqrt{2 + x} - 2]}{\frac{d}{d x} [x - 2]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 2} \frac{\frac{d}{d x} [\sqrt{2 + x} - 2]}{\frac{d}{d x} [x - 2]}$

Step 3.2

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

$lim_{x \to 2} \frac{\frac{d}{d x} [\sqrt{2 + x}] + \frac{d}{d x} [- 2]}{\frac{d}{d x} [x - 2]}$

Step 3.3

Evaluate $\frac{d}{d x} [\sqrt{2 + x}]$ .

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Step 3.3.1

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

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

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = 2 + x$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u$ as $2 + x$ .

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

Step 3.3.2.2

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

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

Step 3.3.2.3

Replace all occurrences of $u$ with $2 + x$ .

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

Step 3.3.3

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

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

Step 3.3.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 3.3.5

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

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

Step 3.3.6

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

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

Step 3.3.7

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

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

Step 3.3.8

Combine the numerators over the common denominator.

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

Step 3.3.9

Simplify the numerator.

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Step 3.3.9.1

Multiply $- 1$ by $2$ .

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

Step 3.3.9.2

Subtract $2$ from $1$ .

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

Step 3.3.10

Move the negative in front of the fraction.

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

Step 3.3.11

Add $0$ and $1$ .

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

Step 3.3.12

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

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

Step 3.3.13

Multiply $\frac{{(2 + x)}^{- \frac{1}{2}}}{2}$ by $1$ .

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

Step 3.3.14

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

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

Step 3.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 3.5

Add $\frac{1}{2 {(2 + x)}^{\frac{1}{2}}}$ and $0$ .

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

Step 3.6

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

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

Step 3.7

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

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

Step 3.8

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 3.9

Add $1$ and $0$ .

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

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

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

Step 6

Evaluate the limit.

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Step 6.1

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

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

Step 6.2

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 6.3

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $2$ .

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

Step 6.4

Evaluate the limit of $1$ which is constant as $x$ approaches $2$ .

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

Step 6.5

Move the limit under the radical sign.

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

Step 6.6

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 6.7

Evaluate the limit of $2$ which is constant as $x$ approaches $2$ .

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

Step 7

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{2 + 2}}$

Step 8

Simplify the answer.

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Step 8.1

Simplify the denominator.

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Step 8.1.1

Add $2$ and $2$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{4}}$

Step 8.1.2

Rewrite $4$ as $2^{2}$ .

$\frac{1}{2} \cdot \frac{1}{\sqrt{2^{2}}}$

Step 8.1.3

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

$\frac{1}{2} \cdot \frac{1}{2}$

Step 8.2

Multiply $\frac{1}{2} \cdot \frac{1}{2}$ .

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Step 8.2.1

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

$\frac{1}{2 \cdot 2}$

Step 8.2.2

Multiply $2$ by $2$ .

$\frac{1}{4}$