Calculus Examples

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

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.

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

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 $- 1$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Move the limit under the radical sign.

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

Step 1.1.2.4

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

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

Step 1.1.2.5

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

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

Step 1.1.2.6

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

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

Step 1.1.2.7

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

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

Step 1.1.2.8

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

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Step 1.1.2.8.1

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

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

Step 1.1.2.8.2

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

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

Step 1.1.2.9

Simplify the answer.

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Step 1.1.2.9.1

Simplify each term.

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Step 1.1.2.9.1.1

Raise $- 1$ to the power of $2$ .

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

Step 1.1.2.9.1.2

Multiply $2$ by $- 1$ .

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

Step 1.1.2.9.1.3

Subtract $2$ from $1$ .

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

Step 1.1.2.9.1.4

Add $- 1$ and $2$ .

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

Step 1.1.2.9.1.5

Any root of $1$ is $1$ .

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

Step 1.1.2.9.1.6

Multiply $- 1$ by $1$ .

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

Step 1.1.2.9.2

Subtract $1$ from $1$ .

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

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 $- 1$ .

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

Step 1.1.3.1.2

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

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

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

Subtract $1$ from $1$ .

$\frac{0}{0}$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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Step 1.3.4.1

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

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

Step 1.3.4.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- {(x^{2} + 2 x + 2)}^{\frac{1}{2}}$ with respect to $x$ is $- \frac{d}{d x} [{(x^{2} + 2 x + 2)}^{\frac{1}{2}}]$ .

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

Step 1.3.4.3

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

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Step 1.3.4.3.1

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

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

Step 1.3.4.3.2

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

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

Step 1.3.4.3.3

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

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

Step 1.3.4.4

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

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

Step 1.3.4.5

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

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

Step 1.3.4.6

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 1.3.4.7

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

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

Step 1.3.4.8

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

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

Step 1.3.4.9

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

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

Step 1.3.4.10

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

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

Step 1.3.4.11

Combine the numerators over the common denominator.

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

Step 1.3.4.12

Simplify the numerator.

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Step 1.3.4.12.1

Multiply $- 1$ by $2$ .

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

Step 1.3.4.12.2

Subtract $2$ from $1$ .

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

Step 1.3.4.13

Move the negative in front of the fraction.

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

Step 1.3.4.14

Multiply $2$ by $1$ .

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

Step 1.3.4.15

Add $2 x + 2$ and $0$ .

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

Step 1.3.4.16

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

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

Step 1.3.4.17

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

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

Step 1.3.5

Simplify.

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Step 1.3.5.1

Subtract $\frac{1}{2 {(x^{2} + 2 x + 2)}^{\frac{1}{2}}} (2 x + 2)$ from $0$ .

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

Step 1.3.5.2

Reorder the factors of $- \frac{1}{2 {(x^{2} + 2 x + 2)}^{\frac{1}{2}}} (2 x + 2)$ .

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

Step 1.3.5.3

Apply the distributive property.

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

Step 1.3.5.4

Multiply $2$ by $- 1$ .

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

Step 1.3.5.5

Multiply $- 1$ by $2$ .

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

Step 1.3.5.6

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

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

Step 1.3.5.7

Factor $2$ out of $- 2 x$ .

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

Step 1.3.5.8

Factor $2$ out of $- 2$ .

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

Step 1.3.5.9

Factor $2$ out of $2 (- x) + 2 (- 1)$ .

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

Step 1.3.5.10

Cancel the common factors.

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Step 1.3.5.10.1

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

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

Step 1.3.5.10.2

Cancel the common factor.

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

Step 1.3.5.10.3

Rewrite the expression.

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

Step 1.3.5.11

Factor $- 1$ out of $- x$ .

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

Step 1.3.5.12

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 1.3.5.13

Factor $- 1$ out of $- (x) - 1 (1)$ .

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

Step 1.3.5.14

Rewrite $- (x + 1)$ as $- 1 (x + 1)$ .

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

Step 1.3.5.15

Move the negative in front of the fraction.

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

Step 1.3.6

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

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

Step 1.3.7

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

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

Step 1.3.8

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

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

Step 1.3.9

Add $0$ and $1$ .

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

Step 1.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5

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

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

Step 1.6

Multiply $- 1$ by $1$ .

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

Step 2

Evaluate the limit.

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Step 2.1

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Move the limit under the radical sign.

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

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

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

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 $x$ .

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

Step 3.2

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

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

Step 3.3

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

$- \frac{- 1 + 1}{\sqrt{{(- 1)}^{2} + 2 \cdot - 1 + 2}}$

Step 4

Simplify the answer.

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Step 4.1

Add $- 1$ and $1$ .

$- \frac{0}{\sqrt{{(- 1)}^{2} + 2 \cdot - 1 + 2}}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Raise $- 1$ to the power of $2$ .

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

Step 4.2.2

Multiply $2$ by $- 1$ .

$- \frac{0}{\sqrt{1 - 2 + 2}}$

Step 4.2.3

Subtract $2$ from $1$ .

$- \frac{0}{\sqrt{- 1 + 2}}$

Step 4.2.4

Add $- 1$ and $2$ .

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

Step 4.2.5

Any root of $1$ is $1$ .

$- \frac{0}{1}$

Step 4.3

Divide $0$ by $1$ .

$- 0$

Step 4.4

Multiply $- 1$ by $0$ .

$0$