Calculus Examples

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

Step 1

Combine terms.

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Step 1.1

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

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

Step 1.2

Combine $- 5$ and $\frac{x}{x}$ .

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

Step 1.3

Combine the numerators over the common denominator.

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

Step 2

Simplify the limit argument.

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Step 2.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.2

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

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

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.

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

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

Evaluate the limit.

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Step 3.1.2.1.1

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

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

Step 3.1.2.1.2

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

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

Step 3.1.2.1.3

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

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

Step 3.1.2.2

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

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

Step 3.1.2.3

Simplify the answer.

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Step 3.1.2.3.1

Multiply $- 5$ by $- 1$ .

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

Step 3.1.2.3.2

Add $- 5$ and $5$ .

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

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 $- 1$ .

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

Step 3.1.3.2

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

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

Step 3.1.3.3

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

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

Step 3.1.3.4

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

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Step 3.1.3.4.1

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

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

Step 3.1.3.4.2

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

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

Step 3.1.3.5

Simplify the answer.

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Step 3.1.3.5.1

Add $- 1$ and $1$ .

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

Step 3.1.3.5.2

Multiply $- 1$ by $0$ .

$\frac{0}{0}$

Step 3.1.3.5.3

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

Undefined

$\frac{0}{0}$

Step 3.1.3.6

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.

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

Evaluate $\frac{d}{d x} [- 5 x]$ .

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Step 3.3.4.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

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

Step 3.3.4.2

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

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

Step 3.3.4.3

Multiply $- 5$ by $1$ .

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

Step 3.3.5

Subtract $5$ from $0$ .

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

Step 3.3.6

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 $f (x) = x$ and $g (x) = x + 1$ .

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

Step 3.3.7

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

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

Step 3.3.8

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

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

Step 3.3.9

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

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

Step 3.3.10

Add $1$ and $0$ .

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

Step 3.3.11

Multiply $x$ by $1$ .

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

Step 3.3.12

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

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

Step 3.3.13

Multiply $x + 1$ by $1$ .

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

Step 3.3.14

Add $x$ and $x$ .

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

Step 4

Evaluate the limit.

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Step 4.1

Move the term $- 5$ outside of the limit because it is constant with respect to $x$ .

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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

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

Step 4.5

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

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

Step 4.6

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

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

Step 5

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

$- 5 \frac{1}{2 \cdot - 1 + 1}$

Step 6

Simplify the answer.

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Step 6.1

Simplify the denominator.

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Step 6.1.1

Multiply $2$ by $- 1$ .

$- 5 \frac{1}{- 2 + 1}$

Step 6.1.2

Add $- 2$ and $1$ .

$- 5 (\frac{1}{- 1})$

Step 6.2

Divide $1$ by $- 1$ .

$- 5 \cdot - 1$

Step 6.3

Multiply $- 5$ by $- 1$ .

$5$