Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

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

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

Step 1.2.5

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

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Step 1.2.5.1

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

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

Step 1.2.5.2

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

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

Step 1.2.6

Simplify the answer.

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Step 1.2.6.1

Simplify each term.

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Step 1.2.6.1.1

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

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

Step 1.2.6.1.2

Multiply $2$ by $1$ .

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

Step 1.2.6.1.3

Multiply $2$ by $- 1$ .

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

Step 1.2.6.2

Subtract $2$ from $2$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Move the limit inside the trig function because cosine is continuous.

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

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

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

Step 1.3.7

Simplify terms.

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Step 1.3.7.1

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

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

Step 1.3.7.2

Simplify the answer.

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Step 1.3.7.2.1

Simplify each term.

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Step 1.3.7.2.1.1

Multiply $- 1$ by $1$ .

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

Step 1.3.7.2.1.2

Add $- 1$ and $1$ .

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

Step 1.3.7.2.1.3

The exact value of $\cos (0)$ is $1$ .

$\frac{0}{3 \cdot 1 - 1 \cdot 3}$

Step 1.3.7.2.1.4

Multiply $3$ by $1$ .

$\frac{0}{3 - 1 \cdot 3}$

Step 1.3.7.2.1.5

Multiply $- 1$ by $3$ .

$\frac{0}{3 - 3}$

Step 1.3.7.2.2

Subtract $3$ from $3$ .

$\frac{0}{0}$

Step 1.3.7.2.3

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

Undefined

$\frac{0}{0}$

Step 1.3.7.3

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

Undefined

$\frac{0}{0}$

Step 1.3.8

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

Undefined

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

$lim_{x \to - 1} \frac{2 x^{2} + 2 x}{3 \cos (- 1 - x) - 3} = lim_{x \to - 1} \frac{\frac{d}{d x} [2 x^{2} + 2 x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to - 1} \frac{\frac{d}{d x} [2 x^{2} + 2 x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.2

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

$lim_{x \to - 1} \frac{\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [2 x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.3

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

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Step 3.3.1

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

$lim_{x \to - 1} \frac{2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.3.2

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

$lim_{x \to - 1} \frac{2 (2 x) + \frac{d}{d x} [2 x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.3.3

Multiply $2$ by $2$ .

$lim_{x \to - 1} \frac{4 x + \frac{d}{d x} [2 x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.4

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

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Step 3.4.1

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

$lim_{x \to - 1} \frac{4 x + 2 \frac{d}{d x} [x]}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.4.2

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

$lim_{x \to - 1} \frac{4 x + 2 \cdot 1}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.4.3

Multiply $2$ by $1$ .

$lim_{x \to - 1} \frac{4 x + 2}{\frac{d}{d x} [3 \cos (- 1 - x) - 3]}$

Step 3.5

By the Sum Rule, the derivative of $3 \cos (- 1 - x) - 3$ with respect to $x$ is $\frac{d}{d x} [3 \cos (- 1 - x)] + \frac{d}{d x} [- 3]$ .

$lim_{x \to - 1} \frac{4 x + 2}{\frac{d}{d x} [3 \cos (- 1 - x)] + \frac{d}{d x} [- 3]}$

Step 3.6

Evaluate $\frac{d}{d x} [3 \cos (- 1 - x)]$ .

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Step 3.6.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \cos (- 1 - x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos (- 1 - x)]$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 \frac{d}{d x} [\cos (- 1 - x)] + \frac{d}{d x} [- 3]}$

Step 3.6.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = - 1 - x$ .

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Step 3.6.2.1

To apply the Chain Rule, set $u$ as $- 1 - x$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [- 1 - x]) + \frac{d}{d x} [- 3]}$

Step 3.6.2.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (u) \frac{d}{d x} [- 1 - x]) + \frac{d}{d x} [- 3]}$

Step 3.6.2.3

Replace all occurrences of $u$ with $- 1 - x$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) \frac{d}{d x} [- 1 - x]) + \frac{d}{d x} [- 3]}$

Step 3.6.3

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

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) (\frac{d}{d x} [- 1] + \frac{d}{d x} [- x])) + \frac{d}{d x} [- 3]}$

Step 3.6.4

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

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) (0 + \frac{d}{d x} [- x])) + \frac{d}{d x} [- 3]}$

Step 3.6.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) (0 - \frac{d}{d x} [x])) + \frac{d}{d x} [- 3]}$

Step 3.6.6

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

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) (0 - 1 \cdot 1)) + \frac{d}{d x} [- 3]}$

Step 3.6.7

Multiply $- 1$ by $1$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) (0 - 1)) + \frac{d}{d x} [- 3]}$

Step 3.6.8

Subtract $1$ from $0$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (- \sin (- 1 - x) \cdot - 1) + \frac{d}{d x} [- 3]}$

Step 3.6.9

Multiply $- 1$ by $- 1$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 (1 \sin (- 1 - x)) + \frac{d}{d x} [- 3]}$

Step 3.6.10

Multiply $\sin (- 1 - x)$ by $1$ .

$lim_{x \to - 1} \frac{4 x + 2}{3 \sin (- 1 - x) + \frac{d}{d x} [- 3]}$

Step 3.7

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

$lim_{x \to - 1} \frac{4 x + 2}{3 \sin (- 1 - x) + 0}$

Step 3.8

Add $3 \sin (- 1 - x)$ and $0$ .

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

Step 4

Since the function approaches $- \infty$ from the left and $\infty$ from the right, the limit does not exist.

$Does not exist$