Calculus Examples

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

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.

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

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 $t$ approaches $0$ .

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

Step 1.2.2

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

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

Step 1.2.3

Move the limit into the exponent.

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

Step 1.2.4

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

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

Step 1.2.5

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

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Step 1.2.5.1

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

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

Step 1.2.5.2

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

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

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

Anything raised to $0$ is $1$ .

$\frac{1 + 0 - 1 \cdot 1}{lim_{t \to 0} \ln (t + 1)}$

Step 1.2.6.1.2

Multiply $- 1$ by $1$ .

$\frac{1 + 0 - 1}{lim_{t \to 0} \ln (t + 1)}$

Step 1.2.6.2

Add $1$ and $0$ .

$\frac{1 - 1}{lim_{t \to 0} \ln (t + 1)}$

Step 1.2.6.3

Subtract $1$ from $1$ .

$\frac{0}{lim_{t \to 0} \ln (t + 1)}$

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.

$\frac{0}{\ln (lim_{t \to 0} t + 1)}$

Step 1.3.1.2

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

$\frac{0}{\ln (lim_{t \to 0} t + lim_{t \to 0} 1)}$

Step 1.3.1.3

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

$\frac{0}{\ln (lim_{t \to 0} t + 1)}$

Step 1.3.2

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

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

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Add $0$ and $1$ .

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

Step 1.3.3.2

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

$\frac{0}{0}$

Step 1.3.3.3

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Evaluate $\frac{d}{d t} [- e^{- t}]$ .

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Step 3.5.1

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

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

Step 3.5.2

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

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Step 3.5.2.1

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

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

Step 3.5.2.2

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

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

Step 3.5.2.3

Replace all occurrences of $u$ with $- t$ .

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

Step 3.5.3

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

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

Step 3.5.4

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

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

Step 3.5.5

Multiply $- 1$ by $1$ .

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

Step 3.5.6

Move $- 1$ to the left of $e^{- t}$ .

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

Step 3.5.7

Rewrite $- 1 e^{- t}$ as $- e^{- t}$ .

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

Step 3.5.8

Multiply $- 1$ by $- 1$ .

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

Step 3.5.9

Multiply $e^{- t}$ by $1$ .

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

Step 3.6

Add $0$ and $1$ .

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

Step 3.7

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 $g (t) = t + 1$ .

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Step 3.7.1

To apply the Chain Rule, set $u$ as $t + 1$ .

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

Step 3.7.2

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

$lim_{t \to 0} \frac{1 + e^{- t}}{\frac{1}{u} \frac{d}{d t} [t + 1]}$

Step 3.7.3

Replace all occurrences of $u$ with $t + 1$ .

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

Step 3.8

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

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

Step 3.9

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

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

Step 3.10

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

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

Step 3.11

Add $1$ and $0$ .

$lim_{t \to 0} \frac{1 + e^{- t}}{\frac{1}{t + 1} \cdot 1}$

Step 3.12

Multiply $\frac{1}{t + 1}$ by $1$ .

$lim_{t \to 0} \frac{1 + e^{- t}}{\frac{1}{t + 1}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

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

$lim_{t \to 0} 1 + e^{- t} \cdot lim_{t \to 0} t + 1$

Step 6

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

$(lim_{t \to 0} 1 + lim_{t \to 0} e^{- t}) \cdot lim_{t \to 0} t + 1$

Step 7

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

$(1 + lim_{t \to 0} e^{- t}) \cdot lim_{t \to 0} t + 1$

Step 8

Move the limit into the exponent.

$(1 + e^{lim_{t \to 0} - t}) \cdot lim_{t \to 0} t + 1$

Step 9

Move the term $- 1$ outside of the limit because it is constant with respect to $t$ .

$(1 + e^{- lim_{t \to 0} t}) \cdot lim_{t \to 0} t + 1$

Step 10

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

$(1 + e^{- lim_{t \to 0} t}) \cdot (lim_{t \to 0} t + lim_{t \to 0} 1)$

Step 11

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

$(1 + e^{- lim_{t \to 0} t}) \cdot (lim_{t \to 0} t + 1)$

Step 12

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

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Step 12.1

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

$(1 + e^{0}) \cdot (lim_{t \to 0} t + 1)$

Step 12.2

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

$(1 + e^{0}) \cdot (0 + 1)$

Step 13

Simplify the answer.

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Step 13.1

Anything raised to $0$ is $1$ .

$(1 + 1) \cdot (0 + 1)$

Step 13.2

Add $1$ and $1$ .

$2 \cdot (0 + 1)$

Step 13.3

Add $0$ and $1$ .

$2 \cdot 1$

Step 13.4

Multiply $2$ by $1$ .

$2$