Precalculus Examples

$lim_{x \to - 5} \frac{2 x + 10}{x^{2} + 2 x - 15}$

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.

$\frac{lim_{x \to - 5} 2 x + 10}{lim_{x \to - 5} x^{2} + 2 x - 15}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 5$ .

$\frac{lim_{x \to - 5} 2 x + lim_{x \to - 5} 10}{lim_{x \to - 5} x^{2} + 2 x - 15}$

Step 1.1.2.1.2

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

$\frac{2 lim_{x \to - 5} x + lim_{x \to - 5} 10}{lim_{x \to - 5} x^{2} + 2 x - 15}$

Step 1.1.2.1.3

Evaluate the limit of $10$ which is constant as $x$ approaches $- 5$ .

$\frac{2 lim_{x \to - 5} x + 10}{lim_{x \to - 5} x^{2} + 2 x - 15}$

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $- 5$ for $x$ .

$\frac{2 \cdot - 5 + 10}{lim_{x \to - 5} x^{2} + 2 x - 15}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Multiply $2$ by $- 5$ .

$\frac{- 10 + 10}{lim_{x \to - 5} x^{2} + 2 x - 15}$

Step 1.1.2.3.2

Add $- 10$ and $10$ .

$\frac{0}{lim_{x \to - 5} x^{2} + 2 x - 15}$

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 $- 5$ .

$\frac{0}{lim_{x \to - 5} x^{2} + lim_{x \to - 5} 2 x - lim_{x \to - 5} 15}$

Step 1.1.3.2

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

$\frac{0}{{(lim_{x \to - 5} x)}^{2} + lim_{x \to - 5} 2 x - lim_{x \to - 5} 15}$

Step 1.1.3.3

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

$\frac{0}{{(lim_{x \to - 5} x)}^{2} + 2 lim_{x \to - 5} x - lim_{x \to - 5} 15}$

Step 1.1.3.4

Evaluate the limit of $15$ which is constant as $x$ approaches $- 5$ .

$\frac{0}{{(lim_{x \to - 5} x)}^{2} + 2 lim_{x \to - 5} x - 1 \cdot 15}$

Step 1.1.3.5

Evaluate the limits by plugging in $- 5$ for all occurrences of $x$ .

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Step 1.1.3.5.1

Evaluate the limit of $x$ by plugging in $- 5$ for $x$ .

$\frac{0}{{(- 5)}^{2} + 2 lim_{x \to - 5} x - 1 \cdot 15}$

Step 1.1.3.5.2

Evaluate the limit of $x$ by plugging in $- 5$ for $x$ .

$\frac{0}{{(- 5)}^{2} + 2 \cdot - 5 - 1 \cdot 15}$

Step 1.1.3.6

Simplify the answer.

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Step 1.1.3.6.1

Simplify each term.

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Step 1.1.3.6.1.1

Raise $- 5$ to the power of $2$ .

$\frac{0}{25 + 2 \cdot - 5 - 1 \cdot 15}$

Step 1.1.3.6.1.2

Multiply $2$ by $- 5$ .

$\frac{0}{25 - 10 - 1 \cdot 15}$

Step 1.1.3.6.1.3

Multiply $- 1$ by $15$ .

$\frac{0}{25 - 10 - 15}$

Step 1.1.3.6.2

Subtract $10$ from $25$ .

$\frac{0}{15 - 15}$

Step 1.1.3.6.3

Subtract $15$ from $15$ .

$\frac{0}{0}$

Step 1.1.3.6.4

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

Undefined

$\frac{0}{0}$

Step 1.1.3.7

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.

$lim_{x \to - 5} \frac{2 x + 10}{x^{2} + 2 x - 15} = lim_{x \to - 5} \frac{\frac{d}{d x} [2 x + 10]}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 5} \frac{\frac{d}{d x} [2 x + 10]}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3.2

By the Sum Rule, the derivative of $2 x + 10$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [10]$ .

$lim_{x \to - 5} \frac{\frac{d}{d x} [2 x] + \frac{d}{d x} [10]}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3.3

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

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Step 1.3.3.1

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

$lim_{x \to - 5} \frac{2 \frac{d}{d x} [x] + \frac{d}{d x} [10]}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

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 $n = 1$ .

$lim_{x \to - 5} \frac{2 \cdot 1 + \frac{d}{d x} [10]}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3.3.3

Multiply $2$ by $1$ .

$lim_{x \to - 5} \frac{2 + \frac{d}{d x} [10]}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3.4

Since $10$ is constant with respect to $x$ , the derivative of $10$ with respect to $x$ is $0$ .

$lim_{x \to - 5} \frac{2 + 0}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3.5

Add $2$ and $0$ .

$lim_{x \to - 5} \frac{2}{\frac{d}{d x} [x^{2} + 2 x - 15]}$

Step 1.3.6

By the Sum Rule, the derivative of $x^{2} + 2 x - 15$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 15]$ .

$lim_{x \to - 5} \frac{2}{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 15]}$

Step 1.3.7

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

$lim_{x \to - 5} \frac{2}{2 x + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 15]}$

Step 1.3.8

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

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Step 1.3.8.1

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

$lim_{x \to - 5} \frac{2}{2 x + 2 \frac{d}{d x} [x] + \frac{d}{d x} [- 15]}$

Step 1.3.8.2

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

$lim_{x \to - 5} \frac{2}{2 x + 2 \cdot 1 + \frac{d}{d x} [- 15]}$

Step 1.3.8.3

Multiply $2$ by $1$ .

$lim_{x \to - 5} \frac{2}{2 x + 2 + \frac{d}{d x} [- 15]}$

Step 1.3.9

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15$ with respect to $x$ is $0$ .

$lim_{x \to - 5} \frac{2}{2 x + 2 + 0}$

Step 1.3.10

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

$lim_{x \to - 5} \frac{2}{2 x + 2}$

Step 1.4

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

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Step 1.4.1

Factor $2$ out of $2$ .

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

Step 1.4.2

Cancel the common factors.

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Step 1.4.2.1

Factor $2$ out of $2 x$ .

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

Step 1.4.2.2

Factor $2$ out of $2$ .

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

Step 1.4.2.3

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

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

Step 1.4.2.4

Cancel the common factor.

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

Step 1.4.2.5

Rewrite the expression.

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

Step 2

Evaluate the limit.

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Step 2.1

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $- 5$ .

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

Step 2.2

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

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

Step 2.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 5$ .

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

Step 2.4

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

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

Step 3

Evaluate the limit of $x$ by plugging in $- 5$ for $x$ .

$\frac{1}{- 5 + 1}$

Step 4

Simplify the answer.

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Step 4.1

Add $- 5$ and $1$ .

$\frac{1}{- 4}$

Step 4.2

Move the negative in front of the fraction.

$- \frac{1}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$- \frac{1}{4}$

Decimal Form:

$- 0.25$