Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 1.1.2.2

Move the limit under the radical sign.

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

Step 1.1.2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 1.1.2.4

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

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

Step 1.1.2.5

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

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Step 1.1.2.5.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.1.2.5.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.1.2.6

Simplify the answer.

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Step 1.1.2.6.1

Simplify each term.

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Step 1.1.2.6.1.1

Add $- 1$ and $2$ .

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

Step 1.1.2.6.1.2

Any root of $1$ is $1$ .

$\frac{- 1 + 1}{lim_{x \to - 1} x + 1}$

Step 1.1.2.6.2

Add $- 1$ and $1$ .

$\frac{0}{lim_{x \to - 1} x + 1}$

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

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

$\frac{0}{lim_{x \to - 1} x + lim_{x \to - 1} 1}$

Step 1.1.3.1.2

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

$\frac{0}{lim_{x \to - 1} x + 1}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 1.1.3.3

Add $- 1$ and $1$ .

$\frac{0}{0}$

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.

$lim_{x \to - 1} \frac{x + \sqrt{x + 2}}{x + 1} = lim_{x \to - 1} \frac{\frac{d}{d x} [x + \sqrt{x + 2}]}{\frac{d}{d x} [x + 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

Step 1.3.4.2

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 $g (x) = x + 2$ .

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Step 1.3.4.2.1

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

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

Step 1.3.4.2.2

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

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

Step 1.3.4.2.3

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

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

Step 1.3.4.3

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

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

Step 1.3.4.4

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

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

Step 1.3.4.5

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

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

Step 1.3.4.6

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

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

Step 1.3.4.7

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

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

Step 1.3.4.8

Combine the numerators over the common denominator.

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

Step 1.3.4.9

Simplify the numerator.

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Step 1.3.4.9.1

Multiply $- 1$ by $2$ .

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

Step 1.3.4.9.2

Subtract $2$ from $1$ .

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

Step 1.3.4.10

Move the negative in front of the fraction.

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

Step 1.3.4.11

Add $1$ and $0$ .

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

Step 1.3.4.12

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

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

Step 1.3.4.13

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

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

Step 1.3.4.14

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

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

Step 1.3.5

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

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

Step 1.3.6

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

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

Step 1.3.7

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

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

Step 1.3.8

Add $1$ and $0$ .

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

Step 1.4

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

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

Step 1.5

Combine terms.

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Step 1.5.1

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

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

Step 1.5.2

Combine the numerators over the common denominator.

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

Step 1.6

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

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

Step 2

Evaluate the limit.

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Step 2.1

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

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

Step 2.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 1$ .

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

Step 2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 2.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

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

Step 2.5

Move the limit under the radical sign.

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

Step 2.6

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

Move the limit under the radical sign.

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

Step 2.10

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 1$ .

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

Step 2.11

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

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

Step 3

Evaluate the limits by plugging in $- 1$ for all occurrences of $x$ .

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Step 3.1

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 3.2

Evaluate the limit of $x$ by plugging in $- 1$ for $x$ .

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

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Add $- 1$ and $2$ .

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

Step 4.1.2

Any root of $1$ is $1$ .

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

Step 4.1.3

Multiply $2$ by $1$ .

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

Step 4.1.4

Add $2$ and $1$ .

$\frac{1}{2} \cdot \frac{3}{\sqrt{- 1 + 2}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Add $- 1$ and $2$ .

$\frac{1}{2} \cdot \frac{3}{\sqrt{1}}$

Step 4.2.2

Any root of $1$ is $1$ .

$\frac{1}{2} \cdot \frac{3}{1}$

Step 4.3

Divide $3$ by $1$ .

$\frac{1}{2} \cdot 3$

Step 4.4

Combine $\frac{1}{2}$ and $3$ .

$\frac{3}{2}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$\frac{3}{2}$

Decimal Form:

$1.5$