Calculus Examples

$lim_{x \to - 1} \frac{\frac{1}{x + 3} + \frac{3}{x - 5}}{x + 1}$

Step 1

Combine terms.

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Step 1.1

To write $\frac{1}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$lim_{x \to - 1} \frac{\frac{1}{x + 3} \cdot \frac{x - 5}{x - 5} + \frac{3}{x - 5}}{x + 1}$

Step 1.2

To write $\frac{3}{x - 5}$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$lim_{x \to - 1} \frac{\frac{1}{x + 3} \cdot \frac{x - 5}{x - 5} + \frac{3}{x - 5} \cdot \frac{x + 3}{x + 3}}{x + 1}$

Step 1.3

Write each expression with a common denominator of $(x + 3) (x - 5)$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{1}{x + 3}$ by $\frac{x - 5}{x - 5}$ .

$lim_{x \to - 1} \frac{\frac{x - 5}{(x + 3) (x - 5)} + \frac{3}{x - 5} \cdot \frac{x + 3}{x + 3}}{x + 1}$

Step 1.3.2

Multiply $\frac{3}{x - 5}$ by $\frac{x + 3}{x + 3}$ .

$lim_{x \to - 1} \frac{\frac{x - 5}{(x + 3) (x - 5)} + \frac{3 (x + 3)}{(x - 5) (x + 3)}}{x + 1}$

Step 1.3.3

Reorder the factors of $(x - 5) (x + 3)$ .

$lim_{x \to - 1} \frac{\frac{x - 5}{(x + 3) (x - 5)} + \frac{3 (x + 3)}{(x + 3) (x - 5)}}{x + 1}$

Step 1.4

Combine the numerators over the common denominator.

$lim_{x \to - 1} \frac{\frac{x - 5 + 3 (x + 3)}{(x + 3) (x - 5)}}{x + 1}$

Step 2

Simplify the limit argument.

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Step 2.1

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to - 1} \frac{x - 5 + 3 (x + 3)}{(x + 3) (x - 5)} \cdot \frac{1}{x + 1}$

Step 2.2

Multiply $\frac{x - 5 + 3 (x + 3)}{(x + 3) (x - 5)}$ by $\frac{1}{x + 1}$ .

$lim_{x \to - 1} \frac{x - 5 + 3 (x + 3)}{(x + 3) (x - 5) (x + 1)}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

$\frac{lim_{x \to - 1} x - 5 + 3 (x + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

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

$\frac{lim_{x \to - 1} x - lim_{x \to - 1} 5 + lim_{x \to - 1} 3 (x + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.2

Evaluate the limit of $5$ which is constant as $x$ approaches $- 1$ .

$\frac{lim_{x \to - 1} x - 1 \cdot 5 + lim_{x \to - 1} 3 (x + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.3

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$\frac{lim_{x \to - 1} x - 1 \cdot 5 + 3 lim_{x \to - 1} x + 3}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.4

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

$\frac{lim_{x \to - 1} x - 1 \cdot 5 + 3 (lim_{x \to - 1} x + lim_{x \to - 1} 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.5

Evaluate the limit of $3$ which is constant as $x$ approaches $- 1$ .

$\frac{lim_{x \to - 1} x - 1 \cdot 5 + 3 (lim_{x \to - 1} x + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.6

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

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Step 3.1.2.6.1

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

$\frac{- 1 - 1 \cdot 5 + 3 (lim_{x \to - 1} x + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.6.2

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

$\frac{- 1 - 1 \cdot 5 + 3 (- 1 + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.7

Simplify the answer.

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Step 3.1.2.7.1

Simplify each term.

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Step 3.1.2.7.1.1

Multiply $- 1$ by $5$ .

$\frac{- 1 - 5 + 3 (- 1 + 3)}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.7.1.2

Add $- 1$ and $3$ .

$\frac{- 1 - 5 + 3 \cdot 2}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.7.1.3

Multiply $3$ by $2$ .

$\frac{- 1 - 5 + 6}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.7.2

Subtract $5$ from $- 1$ .

$\frac{- 6 + 6}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.2.7.3

Add $- 6$ and $6$ .

$\frac{0}{lim_{x \to - 1} (x + 3) (x - 5) (x + 1)}$

Step 3.1.3

Evaluate the limit of the denominator.

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Step 3.1.3.1

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

$\frac{0}{lim_{x \to - 1} x + 3 \cdot lim_{x \to - 1} x - 5 \cdot lim_{x \to - 1} x + 1}$

Step 3.1.3.2

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

$\frac{0}{(lim_{x \to - 1} x + lim_{x \to - 1} 3) \cdot lim_{x \to - 1} x - 5 \cdot lim_{x \to - 1} x + 1}$

Step 3.1.3.3

Evaluate the limit of $3$ which is constant as $x$ approaches $- 1$ .

$\frac{0}{(lim_{x \to - 1} x + 3) \cdot lim_{x \to - 1} x - 5 \cdot lim_{x \to - 1} x + 1}$

Step 3.1.3.4

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

$\frac{0}{(lim_{x \to - 1} x + 3) \cdot (lim_{x \to - 1} x - lim_{x \to - 1} 5) \cdot lim_{x \to - 1} x + 1}$

Step 3.1.3.5

Evaluate the limit of $5$ which is constant as $x$ approaches $- 1$ .

$\frac{0}{(lim_{x \to - 1} x + 3) \cdot (lim_{x \to - 1} x - 1 \cdot 5) \cdot lim_{x \to - 1} x + 1}$

Step 3.1.3.6

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

$\frac{0}{(lim_{x \to - 1} x + 3) \cdot (lim_{x \to - 1} x - 1 \cdot 5) \cdot (lim_{x \to - 1} x + lim_{x \to - 1} 1)}$

Step 3.1.3.7

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

$\frac{0}{(lim_{x \to - 1} x + 3) \cdot (lim_{x \to - 1} x - 1 \cdot 5) \cdot (lim_{x \to - 1} x + 1)}$

Step 3.1.3.8

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

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Step 3.1.3.8.1

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

$\frac{0}{(- 1 + 3) \cdot (lim_{x \to - 1} x - 1 \cdot 5) \cdot (lim_{x \to - 1} x + 1)}$

Step 3.1.3.8.2

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

$\frac{0}{(- 1 + 3) \cdot (- 1 - 1 \cdot 5) \cdot (lim_{x \to - 1} x + 1)}$

Step 3.1.3.8.3

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

$\frac{0}{(- 1 + 3) \cdot (- 1 - 1 \cdot 5) \cdot (- 1 + 1)}$

Step 3.1.3.9

Simplify the answer.

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Step 3.1.3.9.1

Add $- 1$ and $3$ .

$\frac{0}{2 \cdot (- 1 - 1 \cdot 5) \cdot (- 1 + 1)}$

Step 3.1.3.9.2

Multiply $- 1$ by $5$ .

$\frac{0}{2 \cdot (- 1 - 5) \cdot (- 1 + 1)}$

Step 3.1.3.9.3

Subtract $5$ from $- 1$ .

$\frac{0}{2 \cdot - 6 \cdot (- 1 + 1)}$

Step 3.1.3.9.4

Multiply $2$ by $- 6$ .

$\frac{0}{- 12 \cdot (- 1 + 1)}$

Step 3.1.3.9.5

Add $- 1$ and $1$ .

$\frac{0}{- 12 \cdot 0}$

Step 3.1.3.9.6

Multiply $- 12$ by $0$ .

$\frac{0}{0}$

Step 3.1.3.9.7

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

Undefined

$\frac{0}{0}$

Step 3.1.3.10

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

Undefined

$\frac{0}{0}$

Step 3.1.4

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

Undefined

$\frac{0}{0}$

Step 3.2

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.

$lim_{x \to - 1} \frac{x - 5 + 3 (x + 3)}{(x + 3) (x - 5) (x + 1)} = lim_{x \to - 1} \frac{\frac{d}{d x} [x - 5 + 3 (x + 3)]}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 1} \frac{\frac{d}{d x} [x - 5 + 3 (x + 3)]}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.2

By the Sum Rule, the derivative of $x - 5 + 3 (x + 3)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5] + \frac{d}{d x} [3 (x + 3)]$ .

$lim_{x \to - 1} \frac{\frac{d}{d x} [x] + \frac{d}{d x} [- 5] + \frac{d}{d x} [3 (x + 3)]}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.3

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

$lim_{x \to - 1} \frac{1 + \frac{d}{d x} [- 5] + \frac{d}{d x} [3 (x + 3)]}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{1 + 0 + \frac{d}{d x} [3 (x + 3)]}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.5

Evaluate $\frac{d}{d x} [3 (x + 3)]$ .

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Step 3.3.5.1

Since $3$ is constant with respect to $x$ , the derivative of $3 (x + 3)$ with respect to $x$ is $3 \frac{d}{d x} [x + 3]$ .

$lim_{x \to - 1} \frac{1 + 0 + 3 \frac{d}{d x} [x + 3]}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.5.2

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$lim_{x \to - 1} \frac{1 + 0 + 3 (\frac{d}{d x} [x] + \frac{d}{d x} [3])}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.5.3

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

$lim_{x \to - 1} \frac{1 + 0 + 3 (1 + \frac{d}{d x} [3])}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.5.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{1 + 0 + 3 (1 + 0)}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.5.5

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{1 + 0 + 3 \cdot 1}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.5.6

Multiply $3$ by $1$ .

$lim_{x \to - 1} \frac{1 + 0 + 3}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.6

Combine terms.

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Step 3.3.6.1

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{1 + 3}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.6.2

Add $1$ and $3$ .

$lim_{x \to - 1} \frac{4}{\frac{d}{d x} [(x + 3) (x - 5) (x + 1)]}$

Step 3.3.7

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = (x + 3) (x - 5)$ and $g (x) = x + 1$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) \frac{d}{d x} [x + 1] + (x + 1) \frac{d}{d x} [(x + 3) (x - 5)]}$

Step 3.3.8

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

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + (x + 1) \frac{d}{d x} [(x + 3) (x - 5)]}$

Step 3.3.9

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

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) (1 + \frac{d}{d x} [1]) + (x + 1) \frac{d}{d x} [(x + 3) (x - 5)]}$

Step 3.3.10

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

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) (1 + 0) + (x + 1) \frac{d}{d x} [(x + 3) (x - 5)]}$

Step 3.3.11

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) \cdot 1 + (x + 1) \frac{d}{d x} [(x + 3) (x - 5)]}$

Step 3.3.12

Multiply $x + 3$ by $1$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) \frac{d}{d x} [(x + 3) (x - 5)]}$

Step 3.3.13

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x + 3$ and $g (x) = x - 5$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) ((x + 3) \frac{d}{d x} [x - 5] + (x - 5) \frac{d}{d x} [x + 3])}$

Step 3.3.14

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) ((x + 3) (\frac{d}{d x} [x] + \frac{d}{d x} [- 5]) + (x - 5) \frac{d}{d x} [x + 3])}$

Step 3.3.15

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

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) ((x + 3) (1 + \frac{d}{d x} [- 5]) + (x - 5) \frac{d}{d x} [x + 3])}$

Step 3.3.16

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) ((x + 3) (1 + 0) + (x - 5) \frac{d}{d x} [x + 3])}$

Step 3.3.17

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) ((x + 3) \cdot 1 + (x - 5) \frac{d}{d x} [x + 3])}$

Step 3.3.18

Multiply $x + 3$ by $1$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (x + 3 + (x - 5) \frac{d}{d x} [x + 3])}$

Step 3.3.19

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (x + 3 + (x - 5) (\frac{d}{d x} [x] + \frac{d}{d x} [3]))}$

Step 3.3.20

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

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (x + 3 + (x - 5) (1 + \frac{d}{d x} [3]))}$

Step 3.3.21

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (x + 3 + (x - 5) (1 + 0))}$

Step 3.3.22

Add $1$ and $0$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (x + 3 + (x - 5) \cdot 1)}$

Step 3.3.23

Multiply $x - 5$ by $1$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (x + 3 + x - 5)}$

Step 3.3.24

Add $x$ and $x$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (2 x + 3 - 5)}$

Step 3.3.25

Subtract $5$ from $3$ .

$lim_{x \to - 1} \frac{4}{(x + 3) (x - 5) + (x + 1) (2 x - 2)}$

Step 3.3.26

Simplify.

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Step 3.3.26.1

Apply the distributive property.

$lim_{x \to - 1} \frac{4}{(x + 3) x + (x + 3) \cdot - 5 + (x + 1) (2 x - 2)}$

Step 3.3.26.2

Apply the distributive property.

$lim_{x \to - 1} \frac{4}{x \cdot x + 3 x + (x + 3) \cdot - 5 + (x + 1) (2 x - 2)}$

Step 3.3.26.3

Apply the distributive property.

$lim_{x \to - 1} \frac{4}{x \cdot x + 3 x + x \cdot - 5 + 3 \cdot - 5 + (x + 1) (2 x - 2)}$

Step 3.3.26.4

Apply the distributive property.

$lim_{x \to - 1} \frac{4}{x \cdot x + 3 x + x \cdot - 5 + 3 \cdot - 5 + (x + 1) (2 x) + (x + 1) \cdot - 2}$

Step 3.3.26.5

Apply the distributive property.

$lim_{x \to - 1} \frac{4}{x \cdot x + 3 x + x \cdot - 5 + 3 \cdot - 5 + x (2 x) + 1 (2 x) + (x + 1) \cdot - 2}$

Step 3.3.26.6

Apply the distributive property.

$lim_{x \to - 1} \frac{4}{x \cdot x + 3 x + x \cdot - 5 + 3 \cdot - 5 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7

Combine terms.

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Step 3.3.26.7.1

Raise $x$ to the power of $1$ .

$lim_{x \to - 1} \frac{4}{x^{1} x + 3 x + x \cdot - 5 + 3 \cdot - 5 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.2

Raise $x$ to the power of $1$ .

$lim_{x \to - 1} \frac{4}{x^{1} x^{1} + 3 x + x \cdot - 5 + 3 \cdot - 5 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$lim_{x \to - 1} \frac{4}{x^{1 + 1} + 3 x + x \cdot - 5 + 3 \cdot - 5 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.4

Add $1$ and $1$ .

$lim_{x \to - 1} \frac{4}{x^{2} + 3 x + x \cdot - 5 + 3 \cdot - 5 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.5

Move $- 5$ to the left of $x$ .

$lim_{x \to - 1} \frac{4}{x^{2} + 3 x - 5 \cdot x + 3 \cdot - 5 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.6

Multiply $3$ by $- 5$ .

$lim_{x \to - 1} \frac{4}{x^{2} + 3 x - 5 x - 15 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.7

Subtract $5 x$ from $3 x$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + x (2 x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.8

Raise $x$ to the power of $1$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 (x^{1} x) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.9

Raise $x$ to the power of $1$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 (x^{1} x^{1}) + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.10

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{1 + 1} + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.11

Add $1$ and $1$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{2} + 1 (2 x) + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.12

Multiply $2$ by $1$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{2} + 2 x + x \cdot - 2 + 1 \cdot - 2}$

Step 3.3.26.7.13

Move $- 2$ to the left of $x$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{2} + 2 x - 2 \cdot x + 1 \cdot - 2}$

Step 3.3.26.7.14

Multiply $- 2$ by $1$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{2} + 2 x - 2 x - 2}$

Step 3.3.26.7.15

Subtract $2 x$ from $2 x$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{2} + 0 - 2}$

Step 3.3.26.7.16

Add $2 x^{2}$ and $0$ .

$lim_{x \to - 1} \frac{4}{x^{2} - 2 x - 15 + 2 x^{2} - 2}$

Step 3.3.26.7.17

Add $x^{2}$ and $2 x^{2}$ .

$lim_{x \to - 1} \frac{4}{3 x^{2} - 2 x - 15 - 2}$

Step 3.3.26.7.18

Subtract $2$ from $- 15$ .

$lim_{x \to - 1} \frac{4}{3 x^{2} - 2 x - 17}$

Step 4

Evaluate the limit.

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Step 4.1

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$4 lim_{x \to - 1} \frac{1}{3 x^{2} - 2 x - 17}$

Step 4.2

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

$4 \frac{lim_{x \to - 1} 1}{lim_{x \to - 1} 3 x^{2} - 2 x - 17}$

Step 4.3

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

$4 \frac{1}{lim_{x \to - 1} 3 x^{2} - 2 x - 17}$

Step 4.4

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

$4 \frac{1}{lim_{x \to - 1} 3 x^{2} - lim_{x \to - 1} 2 x - lim_{x \to - 1} 17}$

Step 4.5

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$4 \frac{1}{3 lim_{x \to - 1} x^{2} - lim_{x \to - 1} 2 x - lim_{x \to - 1} 17}$

Step 4.6

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

$4 \frac{1}{3 {(lim_{x \to - 1} x)}^{2} - lim_{x \to - 1} 2 x - lim_{x \to - 1} 17}$

Step 4.7

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

$4 \frac{1}{3 {(lim_{x \to - 1} x)}^{2} - 2 lim_{x \to - 1} x - lim_{x \to - 1} 17}$

Step 4.8

Evaluate the limit of $17$ which is constant as $x$ approaches $- 1$ .

$4 \frac{1}{3 {(lim_{x \to - 1} x)}^{2} - 2 lim_{x \to - 1} x - 1 \cdot 17}$

Step 5

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

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Step 5.1

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

$4 \frac{1}{3 {(- 1)}^{2} - 2 lim_{x \to - 1} x - 1 \cdot 17}$

Step 5.2

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

$4 \frac{1}{3 {(- 1)}^{2} - 2 \cdot - 1 - 1 \cdot 17}$

Step 6

Simplify the answer.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

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

$4 \frac{1}{3 \cdot 1 - 2 \cdot - 1 - 1 \cdot 17}$

Step 6.1.2

Multiply $3$ by $1$ .

$4 \frac{1}{3 - 2 \cdot - 1 - 1 \cdot 17}$

Step 6.1.3

Multiply $- 2$ by $- 1$ .

$4 \frac{1}{3 + 2 - 1 \cdot 17}$

Step 6.1.4

Multiply $- 1$ by $17$ .

$4 \frac{1}{3 + 2 - 17}$

Step 6.1.5

Add $3$ and $2$ .

$4 \frac{1}{5 - 17}$

Step 6.1.6

Subtract $17$ from $5$ .

$4 (\frac{1}{- 12})$

Step 6.2

Cancel the common factor of $4$ .

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Step 6.2.1

Factor $4$ out of $- 12$ .

$4 \frac{1}{4 (- 3)}$

Step 6.2.2

Cancel the common factor.

$4 \frac{1}{4 \cdot - 3}$

Step 6.2.3

Rewrite the expression.

$\frac{1}{- 3}$

Step 6.3

Move the negative in front of the fraction.

$- \frac{1}{3}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{3}$

Decimal Form:

$- 0.\overline{3}$