Calculus Examples

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

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.

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

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

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 1.2.1.2

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

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

Step 1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 1.2.1.4

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

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

Step 1.2.1.5

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

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

Step 1.2.1.6

Evaluate the limit of $1$ which is constant as $x$ approaches $2$ .

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

Step 1.2.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

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 $- 1 \cdot 2 \cdot 2$ .

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Step 1.2.3.1.1.1

Multiply $- 1$ by $2$ .

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

Step 1.2.3.1.1.2

Multiply $- 2$ by $2$ .

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

Step 1.2.3.1.2

Subtract $4$ from $4$ .

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

Step 1.2.3.1.3

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

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

Step 1.2.3.1.4

Multiply $- 1$ by $1$ .

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

Step 1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 2} \ln (2 x - 3)}$

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

Move the limit inside the logarithm.

$\frac{0}{\ln (lim_{x \to 2} 2 x - 3)}$

Step 1.3.1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{0}{\ln (lim_{x \to 2} 2 x - lim_{x \to 2} 3)}$

Step 1.3.1.3

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

$\frac{0}{\ln (2 lim_{x \to 2} x - lim_{x \to 2} 3)}$

Step 1.3.1.4

Evaluate the limit of $3$ which is constant as $x$ approaches $2$ .

$\frac{0}{\ln (2 lim_{x \to 2} x - 1 \cdot 3)}$

Step 1.3.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{\ln (2 \cdot 2 - 1 \cdot 3)}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Multiply $2$ by $2$ .

$\frac{0}{\ln (4 - 1 \cdot 3)}$

Step 1.3.3.1.2

Multiply $- 1$ by $3$ .

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

Step 1.3.3.2

Subtract $3$ from $4$ .

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

Step 1.3.3.3

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

$\frac{0}{0}$

Step 1.3.3.4

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

Undefined

$\frac{0}{0}$

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

By the Sum Rule, the derivative of $\cos (4 - 2 x) - 1$ with respect to $x$ is $\frac{d}{d x} [\cos (4 - 2 x)] + \frac{d}{d x} [- 1]$ .

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

Step 3.3

Evaluate $\frac{d}{d x} [\cos (4 - 2 x)]$ .

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Step 3.3.1

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 $g (x) = 4 - 2 x$ .

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Step 3.3.1.1

To apply the Chain Rule, set $u$ as $4 - 2 x$ .

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

Step 3.3.1.2

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

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

Step 3.3.1.3

Replace all occurrences of $u$ with $4 - 2 x$ .

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

Step 3.3.2

By the Sum Rule, the derivative of $4 - 2 x$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- 2 x]$ .

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

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

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

Step 3.3.6

Multiply $- 2$ by $1$ .

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

Step 3.3.7

Subtract $2$ from $0$ .

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

Step 3.3.8

Multiply $- 2$ by $- 1$ .

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

Step 3.4

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

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

Step 3.5

Add $2 \sin (4 - 2 x)$ and $0$ .

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

Step 3.6

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 $g (x) = 2 x - 3$ .

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Step 3.6.1

To apply the Chain Rule, set $u$ as $2 x - 3$ .

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

Step 3.6.2

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

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

Step 3.6.3

Replace all occurrences of $u$ with $2 x - 3$ .

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

Step 3.7

By the Sum Rule, the derivative of $2 x - 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 3]$ .

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

Step 3.8

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

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

Step 3.9

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

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

Step 3.10

Multiply $2$ by $1$ .

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

Step 3.11

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

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

Step 3.12

Add $2$ and $0$ .

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

Step 3.13

Combine $\frac{1}{2 x - 3}$ and $2$ .

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

Combine factors.

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Step 5.1

Combine $\frac{2 x - 3}{2}$ and $2$ .

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

Step 5.2

Combine $\frac{(2 x - 3) \cdot 2}{2}$ and $\sin (4 - 2 x)$ .

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

Step 6

Cancel the common factor of $2$ .

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Step 6.1

Cancel the common factor.

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

Step 6.2

Divide $(2 x - 3) \sin (4 - 2 x)$ by $1$ .

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

Step 7

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 8

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 9

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

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

Step 10

Evaluate the limit of $3$ which is constant as $x$ approaches $2$ .

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

Step 11

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

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

Step 12

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

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

Step 13

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

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

Step 14

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

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

Step 15

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

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Step 15.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 15.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 16

Simplify the answer.

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Step 16.1

Simplify each term.

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Step 16.1.1

Multiply $2$ by $2$ .

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

Step 16.1.2

Multiply $- 1$ by $3$ .

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

Step 16.2

Subtract $3$ from $4$ .

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

Step 16.3

Multiply $\sin (4 - 1 \cdot 2 \cdot 2)$ by $1$ .

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

Step 16.4

Multiply $- 1 \cdot 2 \cdot 2$ .

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Step 16.4.1

Multiply $- 1$ by $2$ .

$\sin (4 - 2 \cdot 2)$

Step 16.4.2

Multiply $- 2$ by $2$ .

$\sin (4 - 4)$

Step 16.5

Subtract $4$ from $4$ .

$\sin (0)$

Step 16.6

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

$0$