Calculus Examples

$lim_{x \to 3} \frac{\sin (2 x - 6)}{\ln (4 - x)}$

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.

$\frac{lim_{x \to 3} \sin (2 x - 6)}{lim_{x \to 3} \ln (4 - x)}$

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 limit inside the trig function because sine is continuous.

$\frac{\sin (lim_{x \to 3} 2 x - 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.1.2

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

$\frac{\sin (lim_{x \to 3} 2 x - lim_{x \to 3} 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.1.3

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

$\frac{\sin (2 lim_{x \to 3} x - lim_{x \to 3} 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.1.4

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

$\frac{\sin (2 lim_{x \to 3} x - 1 \cdot 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.2

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

$\frac{\sin (2 \cdot 3 - 1 \cdot 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Multiply $2$ by $3$ .

$\frac{\sin (6 - 1 \cdot 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.3.1.2

Multiply $- 1$ by $6$ .

$\frac{\sin (6 - 6)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.3.2

Subtract $6$ from $6$ .

$\frac{\sin (0)}{lim_{x \to 3} \ln (4 - x)}$

Step 1.2.3.3

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

$\frac{0}{lim_{x \to 3} \ln (4 - x)}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Move the limit inside the logarithm.

$\frac{0}{\ln (lim_{x \to 3} 4 - x)}$

Step 1.3.2

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

$\frac{0}{\ln (lim_{x \to 3} 4 - lim_{x \to 3} x)}$

Step 1.3.3

Evaluate the limit of $4$ which is constant as $x$ approaches $3$ .

$\frac{0}{\ln (4 - lim_{x \to 3} x)}$

Step 1.3.4

Simplify terms.

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Step 1.3.4.1

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

$\frac{0}{\ln (4 - 3)}$

Step 1.3.4.2

Simplify the answer.

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Step 1.3.4.2.1

Subtract $3$ from $4$ .

$\frac{0}{\ln (1)}$

Step 1.3.4.2.2

The natural logarithm of $1$ is $0$ .

$\frac{0}{0}$

Step 1.3.4.2.3

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

Undefined

$\frac{0}{0}$

Step 1.3.4.3

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

Undefined

$\frac{0}{0}$

Step 1.3.5

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.

$lim_{x \to 3} \frac{\sin (2 x - 6)}{\ln (4 - x)} = lim_{x \to 3} \frac{\frac{d}{d x} [\sin (2 x - 6)]}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 3} \frac{\frac{d}{d x} [\sin (2 x - 6)]}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.2

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

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Step 3.2.1

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

$lim_{x \to 3} \frac{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x - 6]}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.2.2

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

$lim_{x \to 3} \frac{\cos (u) \frac{d}{d x} [2 x - 6]}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.2.3

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

$lim_{x \to 3} \frac{\cos (2 x - 6) \frac{d}{d x} [2 x - 6]}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.3

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

$lim_{x \to 3} \frac{\cos (2 x - 6) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 6])}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.4

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

$lim_{x \to 3} \frac{\cos (2 x - 6) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 6])}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.5

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

$lim_{x \to 3} \frac{\cos (2 x - 6) (2 \cdot 1 + \frac{d}{d x} [- 6])}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.6

Multiply $2$ by $1$ .

$lim_{x \to 3} \frac{\cos (2 x - 6) (2 + \frac{d}{d x} [- 6])}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.7

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

$lim_{x \to 3} \frac{\cos (2 x - 6) (2 + 0)}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.8

Add $2$ and $0$ .

$lim_{x \to 3} \frac{\cos (2 x - 6) \cdot 2}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.9

Move $2$ to the left of $\cos (2 x - 6)$ .

$lim_{x \to 3} \frac{2 \cdot \cos (2 x - 6)}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.10

Multiply $2$ by $\cos (2 x - 6)$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{d}{d x} [\ln (4 - x)]}$

Step 3.11

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 4 - x$ .

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Step 3.11.1

To apply the Chain Rule, set $u$ as $4 - x$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [4 - x]}$

Step 3.11.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{u} \frac{d}{d x} [4 - x]}$

Step 3.11.3

Replace all occurrences of $u$ with $4 - x$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} \frac{d}{d x} [4 - x]}$

Step 3.12

By the Sum Rule, the derivative of $4 - x$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x]$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} (\frac{d}{d x} [4] + \frac{d}{d x} [- x])}$

Step 3.13

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} (0 + \frac{d}{d x} [- x])}$

Step 3.14

Add $0$ and $\frac{d}{d x} [- x]$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} \frac{d}{d x} [- x]}$

Step 3.15

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

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} (- \frac{d}{d x} [x])}$

Step 3.16

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

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} (- 1 \cdot 1)}$

Step 3.17

Multiply $- 1$ by $1$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{1}{4 - x} \cdot - 1}$

Step 3.18

Combine $\frac{1}{4 - x}$ and $- 1$ .

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{\frac{- 1}{4 - x}}$

Step 3.19

Move the negative in front of the fraction.

$lim_{x \to 3} \frac{2 \cos (2 x - 6)}{- \frac{1}{4 - x}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to 3} 2 \cos (2 x - 6) (- (4 - x))$

Step 5

Multiply $- 1$ by $2$ .

$lim_{x \to 3} - 2 \cos (2 x - 6) (4 - x)$

Step 6

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

$- 2 lim_{x \to 3} \cos (2 x - 6) (4 - x)$

Step 7

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

$- 2 lim_{x \to 3} \cos (2 x - 6) \cdot lim_{x \to 3} 4 - x$

Step 8

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

$- 2 \cos (lim_{x \to 3} 2 x - 6) \cdot lim_{x \to 3} 4 - x$

Step 9

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

$- 2 \cos (lim_{x \to 3} 2 x - lim_{x \to 3} 6) \cdot lim_{x \to 3} 4 - x$

Step 10

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

$- 2 \cos (2 lim_{x \to 3} x - lim_{x \to 3} 6) \cdot lim_{x \to 3} 4 - x$

Step 11

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

$- 2 \cos (2 lim_{x \to 3} x - 1 \cdot 6) \cdot lim_{x \to 3} 4 - x$

Step 12

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

$- 2 \cos (2 lim_{x \to 3} x - 1 \cdot 6) \cdot (lim_{x \to 3} 4 - lim_{x \to 3} x)$

Step 13

Evaluate the limit of $4$ which is constant as $x$ approaches $3$ .

$- 2 \cos (2 lim_{x \to 3} x - 1 \cdot 6) \cdot (4 - lim_{x \to 3} x)$

Step 14

Evaluate the limits by plugging in $3$ for all occurrences of $x$ .

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Step 14.1

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

$- 2 \cos (2 \cdot 3 - 1 \cdot 6) \cdot (4 - lim_{x \to 3} x)$

Step 14.2

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

$- 2 \cos (2 \cdot 3 - 1 \cdot 6) \cdot (4 - 3)$

Step 15

Simplify the answer.

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Step 15.1

Simplify each term.

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Step 15.1.1

Multiply $2$ by $3$ .

$- 2 \cos (6 - 1 \cdot 6) \cdot (4 - 3)$

Step 15.1.2

Multiply $- 1$ by $6$ .

$- 2 \cos (6 - 6) \cdot (4 - 3)$

Step 15.2

Subtract $6$ from $6$ .

$- 2 \cos (0) \cdot (4 - 3)$

Step 15.3

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

$- 2 \cdot 1 \cdot (4 - 3)$

Step 15.4

Multiply $- 2$ by $1$ .

$- 2 \cdot (4 - 3)$

Step 15.5

Subtract $3$ from $4$ .

$- 2 \cdot 1$

Step 15.6

Multiply $- 2$ by $1$ .

$- 2$