Calculus Examples

$lim_{x \to - 2} \frac{x^{2} + 4 x + 4}{x^{3} + x^{2} - 8 x - 12}$

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.

$\frac{lim_{x \to - 2} x^{2} + 4 x + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

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 $- 2$ .

$\frac{lim_{x \to - 2} x^{2} + lim_{x \to - 2} 4 x + lim_{x \to - 2} 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.2

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

$\frac{{(lim_{x \to - 2} x)}^{2} + lim_{x \to - 2} 4 x + lim_{x \to - 2} 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.3

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

$\frac{{(lim_{x \to - 2} x)}^{2} + 4 lim_{x \to - 2} x + lim_{x \to - 2} 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.4

Evaluate the limit of $4$ which is constant as $x$ approaches $- 2$ .

$\frac{{(lim_{x \to - 2} x)}^{2} + 4 lim_{x \to - 2} x + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.5

Evaluate the limits by plugging in $- 2$ for all occurrences of $x$ .

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Step 1.1.2.5.1

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

$\frac{{(- 2)}^{2} + 4 lim_{x \to - 2} x + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.5.2

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

$\frac{{(- 2)}^{2} + 4 \cdot - 2 + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

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

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

$\frac{4 + 4 \cdot - 2 + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.6.1.2

Multiply $4$ by $- 2$ .

$\frac{4 - 8 + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.6.2

Subtract $8$ from $4$ .

$\frac{- 4 + 4}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.2.6.3

Add $- 4$ and $4$ .

$\frac{0}{lim_{x \to - 2} x^{3} + x^{2} - 8 x - 12}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

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

$\frac{0}{lim_{x \to - 2} x^{3} + lim_{x \to - 2} x^{2} - lim_{x \to - 2} 8 x - lim_{x \to - 2} 12}$

Step 1.1.3.2

Move the exponent $3$ from $x^{3}$ outside the limit using the Limits Power Rule.

$\frac{0}{{(lim_{x \to - 2} x)}^{3} + lim_{x \to - 2} x^{2} - lim_{x \to - 2} 8 x - lim_{x \to - 2} 12}$

Step 1.1.3.3

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

$\frac{0}{{(lim_{x \to - 2} x)}^{3} + {(lim_{x \to - 2} x)}^{2} - lim_{x \to - 2} 8 x - lim_{x \to - 2} 12}$

Step 1.1.3.4

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{{(lim_{x \to - 2} x)}^{3} + {(lim_{x \to - 2} x)}^{2} - 8 lim_{x \to - 2} x - lim_{x \to - 2} 12}$

Step 1.1.3.5

Evaluate the limit of $12$ which is constant as $x$ approaches $- 2$ .

$\frac{0}{{(lim_{x \to - 2} x)}^{3} + {(lim_{x \to - 2} x)}^{2} - 8 lim_{x \to - 2} x - 1 \cdot 12}$

Step 1.1.3.6

Evaluate the limits by plugging in $- 2$ for all occurrences of $x$ .

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Step 1.1.3.6.1

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

$\frac{0}{{(- 2)}^{3} + {(lim_{x \to - 2} x)}^{2} - 8 lim_{x \to - 2} x - 1 \cdot 12}$

Step 1.1.3.6.2

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

$\frac{0}{{(- 2)}^{3} + {(- 2)}^{2} - 8 lim_{x \to - 2} x - 1 \cdot 12}$

Step 1.1.3.6.3

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

$\frac{0}{{(- 2)}^{3} + {(- 2)}^{2} - 8 \cdot - 2 - 1 \cdot 12}$

Step 1.1.3.7

Simplify the answer.

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Step 1.1.3.7.1

Simplify each term.

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Step 1.1.3.7.1.1

Raise $- 2$ to the power of $3$ .

$\frac{0}{- 8 + {(- 2)}^{2} - 8 \cdot - 2 - 1 \cdot 12}$

Step 1.1.3.7.1.2

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

$\frac{0}{- 8 + 4 - 8 \cdot - 2 - 1 \cdot 12}$

Step 1.1.3.7.1.3

Multiply $- 8$ by $- 2$ .

$\frac{0}{- 8 + 4 + 16 - 1 \cdot 12}$

Step 1.1.3.7.1.4

Multiply $- 1$ by $12$ .

$\frac{0}{- 8 + 4 + 16 - 12}$

Step 1.1.3.7.2

Add $- 8$ and $4$ .

$\frac{0}{- 4 + 16 - 12}$

Step 1.1.3.7.3

Add $- 4$ and $16$ .

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

Step 1.1.3.7.4

Subtract $12$ from $12$ .

$\frac{0}{0}$

Step 1.1.3.7.5

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

Undefined

$\frac{0}{0}$

Step 1.1.3.8

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.

$lim_{x \to - 2} \frac{x^{2} + 4 x + 4}{x^{3} + x^{2} - 8 x - 12} = lim_{x \to - 2} \frac{\frac{d}{d x} [x^{2} + 4 x + 4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 2} \frac{\frac{d}{d x} [x^{2} + 4 x + 4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} + 4 x + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [4]$ .

$lim_{x \to - 2} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.3

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

$lim_{x \to - 2} \frac{2 x + \frac{d}{d x} [4 x] + \frac{d}{d x} [4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 1.3.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to - 2} \frac{2 x + 4 \frac{d}{d x} [x] + \frac{d}{d x} [4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.4.2

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

$lim_{x \to - 2} \frac{2 x + 4 \cdot 1 + \frac{d}{d x} [4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.4.3

Multiply $4$ by $1$ .

$lim_{x \to - 2} \frac{2 x + 4 + \frac{d}{d x} [4]}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.5

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{2 x + 4 + 0}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.6

Add $2 x + 4$ and $0$ .

$lim_{x \to - 2} \frac{2 x + 4}{\frac{d}{d x} [x^{3} + x^{2} - 8 x - 12]}$

Step 1.3.7

By the Sum Rule, the derivative of $x^{3} + x^{2} - 8 x - 12$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [- 12]$ .

$lim_{x \to - 2} \frac{2 x + 4}{\frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [- 12]}$

Step 1.3.8

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [- 12]}$

Step 1.3.9

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

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [- 12]}$

Step 1.3.10

Evaluate $\frac{d}{d x} [- 8 x]$ .

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Step 1.3.10.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x - 8 \frac{d}{d x} [x] + \frac{d}{d x} [- 12]}$

Step 1.3.10.2

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

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x - 8 \cdot 1 + \frac{d}{d x} [- 12]}$

Step 1.3.10.3

Multiply $- 8$ by $1$ .

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x - 8 + \frac{d}{d x} [- 12]}$

Step 1.3.11

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x - 8 + 0}$

Step 1.3.12

Add $3 x^{2} + 2 x - 8$ and $0$ .

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x - 8}$

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

$\frac{lim_{x \to - 2} 2 x + 4}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

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

$\frac{lim_{x \to - 2} 2 x + lim_{x \to - 2} 4}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.2.1.2

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

$\frac{2 lim_{x \to - 2} x + lim_{x \to - 2} 4}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.2.1.3

Evaluate the limit of $4$ which is constant as $x$ approaches $- 2$ .

$\frac{2 lim_{x \to - 2} x + 4}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.2.2

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

$\frac{2 \cdot - 2 + 4}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Multiply $2$ by $- 2$ .

$\frac{- 4 + 4}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.2.3.2

Add $- 4$ and $4$ .

$\frac{0}{lim_{x \to - 2} 3 x^{2} + 2 x - 8}$

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

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

$\frac{0}{lim_{x \to - 2} 3 x^{2} + lim_{x \to - 2} 2 x - lim_{x \to - 2} 8}$

Step 2.1.3.2

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

$\frac{0}{3 lim_{x \to - 2} x^{2} + lim_{x \to - 2} 2 x - lim_{x \to - 2} 8}$

Step 2.1.3.3

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

$\frac{0}{3 {(lim_{x \to - 2} x)}^{2} + lim_{x \to - 2} 2 x - lim_{x \to - 2} 8}$

Step 2.1.3.4

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

$\frac{0}{3 {(lim_{x \to - 2} x)}^{2} + 2 lim_{x \to - 2} x - lim_{x \to - 2} 8}$

Step 2.1.3.5

Evaluate the limit of $8$ which is constant as $x$ approaches $- 2$ .

$\frac{0}{3 {(lim_{x \to - 2} x)}^{2} + 2 lim_{x \to - 2} x - 1 \cdot 8}$

Step 2.1.3.6

Evaluate the limits by plugging in $- 2$ for all occurrences of $x$ .

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Step 2.1.3.6.1

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

$\frac{0}{3 {(- 2)}^{2} + 2 lim_{x \to - 2} x - 1 \cdot 8}$

Step 2.1.3.6.2

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

$\frac{0}{3 {(- 2)}^{2} + 2 \cdot - 2 - 1 \cdot 8}$

Step 2.1.3.7

Simplify the answer.

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Step 2.1.3.7.1

Simplify each term.

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Step 2.1.3.7.1.1

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

$\frac{0}{3 \cdot 4 + 2 \cdot - 2 - 1 \cdot 8}$

Step 2.1.3.7.1.2

Multiply $3$ by $4$ .

$\frac{0}{12 + 2 \cdot - 2 - 1 \cdot 8}$

Step 2.1.3.7.1.3

Multiply $2$ by $- 2$ .

$\frac{0}{12 - 4 - 1 \cdot 8}$

Step 2.1.3.7.1.4

Multiply $- 1$ by $8$ .

$\frac{0}{12 - 4 - 8}$

Step 2.1.3.7.2

Subtract $4$ from $12$ .

$\frac{0}{8 - 8}$

Step 2.1.3.7.3

Subtract $8$ from $8$ .

$\frac{0}{0}$

Step 2.1.3.7.4

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

Undefined

$\frac{0}{0}$

Step 2.1.3.8

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

Undefined

$\frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{0}{0}$

Step 2.2

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.

$lim_{x \to - 2} \frac{2 x + 4}{3 x^{2} + 2 x - 8} = lim_{x \to - 2} \frac{\frac{d}{d x} [2 x + 4]}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 2} \frac{\frac{d}{d x} [2 x + 4]}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.2

By the Sum Rule, the derivative of $2 x + 4$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [4]$ .

$lim_{x \to - 2} \frac{\frac{d}{d x} [2 x] + \frac{d}{d x} [4]}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.3

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

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Step 2.3.3.1

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

$lim_{x \to - 2} \frac{2 \frac{d}{d x} [x] + \frac{d}{d x} [4]}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.3.2

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

$lim_{x \to - 2} \frac{2 \cdot 1 + \frac{d}{d x} [4]}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.3.3

Multiply $2$ by $1$ .

$lim_{x \to - 2} \frac{2 + \frac{d}{d x} [4]}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{2 + 0}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.5

Add $2$ and $0$ .

$lim_{x \to - 2} \frac{2}{\frac{d}{d x} [3 x^{2} + 2 x - 8]}$

Step 2.3.6

By the Sum Rule, the derivative of $3 x^{2} + 2 x - 8$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 8]$ .

$lim_{x \to - 2} \frac{2}{\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 8]}$

Step 2.3.7

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Step 2.3.7.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$lim_{x \to - 2} \frac{2}{3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 8]}$

Step 2.3.7.2

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

$lim_{x \to - 2} \frac{2}{3 (2 x) + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 8]}$

Step 2.3.7.3

Multiply $2$ by $3$ .

$lim_{x \to - 2} \frac{2}{6 x + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 8]}$

Step 2.3.8

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

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Step 2.3.8.1

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

$lim_{x \to - 2} \frac{2}{6 x + 2 \frac{d}{d x} [x] + \frac{d}{d x} [- 8]}$

Step 2.3.8.2

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

$lim_{x \to - 2} \frac{2}{6 x + 2 \cdot 1 + \frac{d}{d x} [- 8]}$

Step 2.3.8.3

Multiply $2$ by $1$ .

$lim_{x \to - 2} \frac{2}{6 x + 2 + \frac{d}{d x} [- 8]}$

Step 2.3.9

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to - 2} \frac{2}{6 x + 2 + 0}$

Step 2.3.10

Add $6 x + 2$ and $0$ .

$lim_{x \to - 2} \frac{2}{6 x + 2}$

Step 2.4

Cancel the common factor of $2$ and $6 x + 2$ .

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Step 2.4.1

Factor $2$ out of $2$ .

$lim_{x \to - 2} \frac{2 \cdot 1}{6 x + 2}$

Step 2.4.2

Cancel the common factors.

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Step 2.4.2.1

Factor $2$ out of $6 x$ .

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

Step 2.4.2.2

Factor $2$ out of $2$ .

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

Step 2.4.2.3

Factor $2$ out of $2 (3 x) + 2 (1)$ .

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

Step 2.4.2.4

Cancel the common factor.

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

Step 2.4.2.5

Rewrite the expression.

$lim_{x \to - 2} \frac{1}{3 x + 1}$

Step 3

Evaluate the limit.

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Step 3.1

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

$\frac{lim_{x \to - 2} 1}{lim_{x \to - 2} 3 x + 1}$

Step 3.2

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

$\frac{1}{lim_{x \to - 2} 3 x + 1}$

Step 3.3

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

$\frac{1}{lim_{x \to - 2} 3 x + lim_{x \to - 2} 1}$

Step 3.4

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

$\frac{1}{3 lim_{x \to - 2} x + lim_{x \to - 2} 1}$

Step 3.5

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

$\frac{1}{3 lim_{x \to - 2} x + 1}$

Step 4

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

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

Step 5

Simplify the answer.

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Step 5.1

Simplify the denominator.

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Step 5.1.1

Multiply $3$ by $- 2$ .

$\frac{1}{- 6 + 1}$

Step 5.1.2

Add $- 6$ and $1$ .

$\frac{1}{- 5}$

Step 5.2

Move the negative in front of the fraction.

$- \frac{1}{5}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{5}$

Decimal Form:

$- 0.2$