Calculus Examples

$lim_{x \to - 3} \frac{3 \sin (2 x + 6)}{3 + x}$

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.

$\frac{lim_{x \to - 3} 3 \sin (2 x + 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

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

$\frac{3 lim_{x \to - 3} \sin (2 x + 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.1.2

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

$\frac{3 \sin (lim_{x \to - 3} 2 x + 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{3 \sin (lim_{x \to - 3} 2 x + lim_{x \to - 3} 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.1.4

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

$\frac{3 \sin (2 lim_{x \to - 3} x + lim_{x \to - 3} 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.1.5

Evaluate the limit of $6$ which is constant as $x$ approaches $- 3$ .

$\frac{3 \sin (2 lim_{x \to - 3} x + 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{3 \sin (2 \cdot - 3 + 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Multiply $2$ by $- 3$ .

$\frac{3 \sin (- 6 + 6)}{lim_{x \to - 3} 3 + x}$

Step 1.2.3.2

Add $- 6$ and $6$ .

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

Step 1.2.3.3

The exact value of $\sin (0)$ is $0$ .

$\frac{3 \cdot 0}{lim_{x \to - 3} 3 + x}$

Step 1.2.3.4

Multiply $3$ by $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

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

Step 1.3.1.2

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

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

Step 1.3.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

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

Step 1.3.3

Subtract $3$ from $3$ .

$\frac{0}{0}$

Step 1.3.4

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.

$lim_{x \to - 3} \frac{3 \sin (2 x + 6)}{3 + x} = lim_{x \to - 3} \frac{\frac{d}{d x} [3 \sin (2 x + 6)]}{\frac{d}{d x} [3 + x]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to - 3} \frac{\frac{d}{d x} [3 \sin (2 x + 6)]}{\frac{d}{d x} [3 + x]}$

Step 3.2

Since $3$ is constant with respect to $x$ , the derivative of $3 \sin (2 x + 6)$ with respect to $x$ is $3 \frac{d}{d x} [\sin (2 x + 6)]$ .

$lim_{x \to - 3} \frac{3 \frac{d}{d x} [\sin (2 x + 6)]}{\frac{d}{d x} [3 + x]}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 x + 6$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $2 x + 6$ .

$lim_{x \to - 3} \frac{3 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x + 6])}{\frac{d}{d x} [3 + x]}$

Step 3.3.2

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

$lim_{x \to - 3} \frac{3 (\cos (u) \frac{d}{d x} [2 x + 6])}{\frac{d}{d x} [3 + x]}$

Step 3.3.3

Replace all occurrences of $u$ with $2 x + 6$ .

$lim_{x \to - 3} \frac{3 (\cos (2 x + 6) \frac{d}{d x} [2 x + 6])}{\frac{d}{d x} [3 + x]}$

Step 3.4

Remove parentheses.

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) \frac{d}{d x} [2 x + 6]}{\frac{d}{d x} [3 + x]}$

Step 3.5

By the Sum Rule, the derivative of $2 x + 6$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [6]$ .

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) (\frac{d}{d x} [2 x] + \frac{d}{d x} [6])}{\frac{d}{d x} [3 + x]}$

Step 3.6

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

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) (2 \frac{d}{d x} [x] + \frac{d}{d x} [6])}{\frac{d}{d x} [3 + x]}$

Step 3.7

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

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) (2 \cdot 1 + \frac{d}{d x} [6])}{\frac{d}{d x} [3 + x]}$

Step 3.8

Multiply $2$ by $1$ .

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) (2 + \frac{d}{d x} [6])}{\frac{d}{d x} [3 + x]}$

Step 3.9

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

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) (2 + 0)}{\frac{d}{d x} [3 + x]}$

Step 3.10

Add $2$ and $0$ .

$lim_{x \to - 3} \frac{3 \cos (2 x + 6) \cdot 2}{\frac{d}{d x} [3 + x]}$

Step 3.11

Multiply $2$ by $3$ .

$lim_{x \to - 3} \frac{6 \cos (2 x + 6)}{\frac{d}{d x} [3 + x]}$

Step 3.12

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

$lim_{x \to - 3} \frac{6 \cos (2 x + 6)}{\frac{d}{d x} [3] + \frac{d}{d x} [x]}$

Step 3.13

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

$lim_{x \to - 3} \frac{6 \cos (2 x + 6)}{0 + \frac{d}{d x} [x]}$

Step 3.14

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

$lim_{x \to - 3} \frac{6 \cos (2 x + 6)}{0 + 1}$

Step 3.15

Add $0$ and $1$ .

$lim_{x \to - 3} \frac{6 \cos (2 x + 6)}{1}$

Step 4

Evaluate the limit.

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Step 4.1

Divide $6 \cos (2 x + 6)$ by $1$ .

$lim_{x \to - 3} 6 \cos (2 x + 6)$

Step 4.2

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$6 lim_{x \to - 3} \cos (2 x + 6)$

Step 4.3

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

$6 \cos (lim_{x \to - 3} 2 x + 6)$

Step 4.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$6 \cos (lim_{x \to - 3} 2 x + lim_{x \to - 3} 6)$

Step 4.5

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

$6 \cos (2 lim_{x \to - 3} x + lim_{x \to - 3} 6)$

Step 4.6

Evaluate the limit of $6$ which is constant as $x$ approaches $- 3$ .

$6 \cos (2 lim_{x \to - 3} x + 6)$

Step 5

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$6 \cos (2 \cdot - 3 + 6)$

Step 6

Simplify the answer.

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Step 6.1

Multiply $2$ by $- 3$ .

$6 \cos (- 6 + 6)$

Step 6.2

Add $- 6$ and $6$ .

$6 \cos (0)$

Step 6.3

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

$6 \cdot 1$

Step 6.4

Multiply $6$ by $1$ .

$6$