Precalculus Examples

$lim_{x \to - 6} \frac{2 x^{2} + 9 x - 18}{x + 6}$

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.

$\frac{lim_{x \to - 6} 2 x^{2} + 9 x - 18}{lim_{x \to - 6} x + 6}$

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 $- 6$ .

$\frac{lim_{x \to - 6} 2 x^{2} + lim_{x \to - 6} 9 x - lim_{x \to - 6} 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.2

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

$\frac{2 lim_{x \to - 6} x^{2} + lim_{x \to - 6} 9 x - lim_{x \to - 6} 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.3

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

$\frac{2 {(lim_{x \to - 6} x)}^{2} + lim_{x \to - 6} 9 x - lim_{x \to - 6} 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.4

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$\frac{2 {(lim_{x \to - 6} x)}^{2} + 9 lim_{x \to - 6} x - lim_{x \to - 6} 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.5

Evaluate the limit of $18$ which is constant as $x$ approaches $- 6$ .

$\frac{2 {(lim_{x \to - 6} x)}^{2} + 9 lim_{x \to - 6} x - 1 \cdot 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.6

Evaluate the limits by plugging in $- 6$ for all occurrences of $x$ .

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Step 1.1.2.6.1

Evaluate the limit of $x$ by plugging in $- 6$ for $x$ .

$\frac{2 {(- 6)}^{2} + 9 lim_{x \to - 6} x - 1 \cdot 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.6.2

Evaluate the limit of $x$ by plugging in $- 6$ for $x$ .

$\frac{2 {(- 6)}^{2} + 9 \cdot - 6 - 1 \cdot 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.7

Simplify the answer.

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Step 1.1.2.7.1

Simplify each term.

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Step 1.1.2.7.1.1

Raise $- 6$ to the power of $2$ .

$\frac{2 \cdot 36 + 9 \cdot - 6 - 1 \cdot 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.7.1.2

Multiply $2$ by $36$ .

$\frac{72 + 9 \cdot - 6 - 1 \cdot 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.7.1.3

Multiply $9$ by $- 6$ .

$\frac{72 - 54 - 1 \cdot 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.7.1.4

Multiply $- 1$ by $18$ .

$\frac{72 - 54 - 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.7.2

Subtract $54$ from $72$ .

$\frac{18 - 18}{lim_{x \to - 6} x + 6}$

Step 1.1.2.7.3

Subtract $18$ from $18$ .

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

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 $- 6$ .

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

Step 1.1.3.1.2

Evaluate the limit of $6$ which is constant as $x$ approaches $- 6$ .

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

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $- 6$ for $x$ .

$\frac{0}{- 6 + 6}$

Step 1.1.3.3

Add $- 6$ and $6$ .

$\frac{0}{0}$

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.

$lim_{x \to - 6} \frac{2 x^{2} + 9 x - 18}{x + 6} = lim_{x \to - 6} \frac{\frac{d}{d x} [2 x^{2} + 9 x - 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 6} \frac{\frac{d}{d x} [2 x^{2} + 9 x - 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.2

By the Sum Rule, the derivative of $2 x^{2} + 9 x - 18$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [- 18]$ .

$lim_{x \to - 6} \frac{\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.3

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

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Step 1.3.3.1

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

$lim_{x \to - 6} \frac{2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [9 x] + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.3.2

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

$lim_{x \to - 6} \frac{2 (2 x) + \frac{d}{d x} [9 x] + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.3.3

Multiply $2$ by $2$ .

$lim_{x \to - 6} \frac{4 x + \frac{d}{d x} [9 x] + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [9 x]$ .

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Step 1.3.4.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$lim_{x \to - 6} \frac{4 x + 9 \frac{d}{d x} [x] + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

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 $n = 1$ .

$lim_{x \to - 6} \frac{4 x + 9 \cdot 1 + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.4.3

Multiply $9$ by $1$ .

$lim_{x \to - 6} \frac{4 x + 9 + \frac{d}{d x} [- 18]}{\frac{d}{d x} [x + 6]}$

Step 1.3.5

Since $- 18$ is constant with respect to $x$ , the derivative of $- 18$ with respect to $x$ is $0$ .

$lim_{x \to - 6} \frac{4 x + 9 + 0}{\frac{d}{d x} [x + 6]}$

Step 1.3.6

Add $4 x + 9$ and $0$ .

$lim_{x \to - 6} \frac{4 x + 9}{\frac{d}{d x} [x + 6]}$

Step 1.3.7

By the Sum Rule, the derivative of $x + 6$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [6]$ .

$lim_{x \to - 6} \frac{4 x + 9}{\frac{d}{d x} [x] + \frac{d}{d x} [6]}$

Step 1.3.8

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

$lim_{x \to - 6} \frac{4 x + 9}{1 + \frac{d}{d x} [6]}$

Step 1.3.9

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$lim_{x \to - 6} \frac{4 x + 9}{1 + 0}$

Step 1.3.10

Add $1$ and $0$ .

$lim_{x \to - 6} \frac{4 x + 9}{1}$

Step 1.4

Divide $4 x + 9$ by $1$ .

$lim_{x \to - 6} 4 x + 9$

Step 2

Evaluate the limit.

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Step 2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 6$ .

$lim_{x \to - 6} 4 x + lim_{x \to - 6} 9$

Step 2.2

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

$4 lim_{x \to - 6} x + lim_{x \to - 6} 9$

Step 2.3

Evaluate the limit of $9$ which is constant as $x$ approaches $- 6$ .

$4 lim_{x \to - 6} x + 9$

Step 3

Evaluate the limit of $x$ by plugging in $- 6$ for $x$ .

$4 \cdot - 6 + 9$

Step 4

Simplify the answer.

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Step 4.1

Multiply $4$ by $- 6$ .

$- 24 + 9$

Step 4.2

Add $- 24$ and $9$ .

$- 15$