Calculus Examples

$lim_{t \to 0} \frac{\sin (t)}{\ln (2 e^{t} - 1)}$

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.

$\frac{lim_{t \to 0} \sin (t)}{lim_{t \to 0} \ln (2 e^{t} - 1)}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

$\frac{\sin (lim_{t \to 0} t)}{lim_{t \to 0} \ln (2 e^{t} - 1)}$

Step 1.2.2

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{\sin (0)}{lim_{t \to 0} \ln (2 e^{t} - 1)}$

Step 1.2.3

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

$\frac{0}{lim_{t \to 0} \ln (2 e^{t} - 1)}$

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.

$\frac{0}{\ln (lim_{t \to 0} 2 e^{t} - 1)}$

Step 1.3.1.2

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{0}{\ln (lim_{t \to 0} 2 e^{t} - lim_{t \to 0} 1)}$

Step 1.3.1.3

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

$\frac{0}{\ln (2 lim_{t \to 0} e^{t} - lim_{t \to 0} 1)}$

Step 1.3.1.4

Move the limit into the exponent.

$\frac{0}{\ln (2 e^{lim_{t \to 0} t} - lim_{t \to 0} 1)}$

Step 1.3.1.5

Evaluate the limit of $1$ which is constant as $t$ approaches $0$ .

$\frac{0}{\ln (2 e^{lim_{t \to 0} t} - 1 \cdot 1)}$

Step 1.3.2

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

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

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

Anything raised to $0$ is $1$ .

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

Step 1.3.3.1.2

Multiply $2$ by $1$ .

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

Step 1.3.3.1.3

Multiply $- 1$ by $1$ .

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

Step 1.3.3.2

Subtract $1$ from $2$ .

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

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.

$lim_{t \to 0} \frac{\sin (t)}{\ln (2 e^{t} - 1)} = lim_{t \to 0} \frac{\frac{d}{d t} [\sin (t)]}{\frac{d}{d t} [\ln (2 e^{t} - 1)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{t \to 0} \frac{\frac{d}{d t} [\sin (t)]}{\frac{d}{d t} [\ln (2 e^{t} - 1)]}$

Step 3.2

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

$lim_{t \to 0} \frac{\cos (t)}{\frac{d}{d t} [\ln (2 e^{t} - 1)]}$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \ln (t)$ and $g (t) = 2 e^{t} - 1$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $2 e^{t} - 1$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{d}{d u} [\ln (u)] \frac{d}{d t} [2 e^{t} - 1]}$

Step 3.3.2

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

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{u} \frac{d}{d t} [2 e^{t} - 1]}$

Step 3.3.3

Replace all occurrences of $u$ with $2 e^{t} - 1$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{2 e^{t} - 1} \frac{d}{d t} [2 e^{t} - 1]}$

Step 3.4

By the Sum Rule, the derivative of $2 e^{t} - 1$ with respect to $t$ is $\frac{d}{d t} [2 e^{t}] + \frac{d}{d t} [- 1]$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{2 e^{t} - 1} (\frac{d}{d t} [2 e^{t}] + \frac{d}{d t} [- 1])}$

Step 3.5

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

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{2 e^{t} - 1} (2 \frac{d}{d t} [e^{t}] + \frac{d}{d t} [- 1])}$

Step 3.6

Differentiate using the Exponential Rule which states that $\frac{d}{d t} [a^{t}]$ is $a^{t} \ln (a)$ where $a$ = $e$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{2 e^{t} - 1} (2 e^{t} + \frac{d}{d t} [- 1])}$

Step 3.7

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

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{2 e^{t} - 1} (2 e^{t} + 0)}$

Step 3.8

Add $2 e^{t}$ and $0$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{1}{2 e^{t} - 1} (2 e^{t})}$

Step 3.9

Combine $2$ and $\frac{1}{2 e^{t} - 1}$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{2}{2 e^{t} - 1} e^{t}}$

Step 3.10

Combine $\frac{2}{2 e^{t} - 1}$ and $e^{t}$ .

$lim_{t \to 0} \frac{\cos (t)}{\frac{2 e^{t}}{2 e^{t} - 1}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{t \to 0} \cos (t) \frac{2 e^{t} - 1}{2 e^{t}}$

Step 5

Combine $\cos (t)$ and $\frac{2 e^{t} - 1}{2 e^{t}}$ .

$lim_{t \to 0} \frac{\cos (t) (2 e^{t} - 1)}{2 e^{t}}$

Step 6

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $t$ .

$\frac{1}{2} lim_{t \to 0} \frac{\cos (t) (2 e^{t} - 1)}{e^{t}}$

Step 7

Split the limit using the Limits Quotient Rule on the limit as $t$ approaches $0$ .

$\frac{1}{2} \cdot \frac{lim_{t \to 0} \cos (t) (2 e^{t} - 1)}{lim_{t \to 0} e^{t}}$

Step 8

Split the limit using the Product of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{1}{2} \cdot \frac{lim_{t \to 0} \cos (t) \cdot lim_{t \to 0} 2 e^{t} - 1}{lim_{t \to 0} e^{t}}$

Step 9

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

$\frac{1}{2} \cdot \frac{\cos (lim_{t \to 0} t) \cdot lim_{t \to 0} 2 e^{t} - 1}{lim_{t \to 0} e^{t}}$

Step 10

Split the limit using the Sum of Limits Rule on the limit as $t$ approaches $0$ .

$\frac{1}{2} \cdot \frac{\cos (lim_{t \to 0} t) \cdot (lim_{t \to 0} 2 e^{t} - lim_{t \to 0} 1)}{lim_{t \to 0} e^{t}}$

Step 11

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

$\frac{1}{2} \cdot \frac{\cos (lim_{t \to 0} t) \cdot (2 lim_{t \to 0} e^{t} - lim_{t \to 0} 1)}{lim_{t \to 0} e^{t}}$

Step 12

Move the limit into the exponent.

$\frac{1}{2} \cdot \frac{\cos (lim_{t \to 0} t) \cdot (2 e^{lim_{t \to 0} t} - lim_{t \to 0} 1)}{lim_{t \to 0} e^{t}}$

Step 13

Evaluate the limit of $1$ which is constant as $t$ approaches $0$ .

$\frac{1}{2} \cdot \frac{\cos (lim_{t \to 0} t) \cdot (2 e^{lim_{t \to 0} t} - 1 \cdot 1)}{lim_{t \to 0} e^{t}}$

Step 14

Move the limit into the exponent.

$\frac{1}{2} \cdot \frac{\cos (lim_{t \to 0} t) \cdot (2 e^{lim_{t \to 0} t} - 1 \cdot 1)}{e^{lim_{t \to 0} t}}$

Step 15

Evaluate the limits by plugging in $0$ for all occurrences of $t$ .

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Step 15.1

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{1}{2} \cdot \frac{\cos (0) \cdot (2 e^{lim_{t \to 0} t} - 1 \cdot 1)}{e^{lim_{t \to 0} t}}$

Step 15.2

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{1}{2} \cdot \frac{\cos (0) \cdot (2 e^{0} - 1 \cdot 1)}{e^{lim_{t \to 0} t}}$

Step 15.3

Evaluate the limit of $t$ by plugging in $0$ for $t$ .

$\frac{1}{2} \cdot \frac{\cos (0) \cdot (2 e^{0} - 1 \cdot 1)}{e^{0}}$

Step 16

Simplify the answer.

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Step 16.1

Combine.

$\frac{1 (\cos (0) \cdot (2 e^{0} - 1 \cdot 1))}{2 e^{0}}$

Step 16.2

Multiply $\cos (0)$ by $1$ .

$\frac{\cos (0) (2 e^{0} - 1 \cdot 1)}{2 e^{0}}$

Step 16.3

Anything raised to $0$ is $1$ .

$\frac{\cos (0) (2 e^{0} - 1 \cdot 1)}{2 \cdot 1}$

Step 16.4

Simplify the numerator.

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Step 16.4.1

Anything raised to $0$ is $1$ .

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

Step 16.4.2

Multiply $2$ by $1$ .

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

Step 16.4.3

Multiply $- 1$ by $1$ .

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

Step 16.4.4

Subtract $1$ from $2$ .

$\frac{\cos (0) \cdot 1}{2 \cdot 1}$

Step 16.4.5

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

$\frac{1 \cdot 1}{2 \cdot 1}$

Step 16.4.6

Multiply $1$ by $1$ .

$\frac{1}{2 \cdot 1}$

Step 16.5

Multiply $2$ by $1$ .

$\frac{1}{2}$