Calculus Examples

$lim_{x \to 0} \frac{e^{4 x} - e^{2 x}}{e^{x} - 1}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 0} e^{4 x} - e^{2 x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

$\frac{lim_{x \to 0} e^{4 x} - lim_{x \to 0} e^{2 x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.2

Move the limit into the exponent.

$\frac{e^{lim_{x \to 0} 4 x} - lim_{x \to 0} e^{2 x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.3

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{e^{4 lim_{x \to 0} x} - lim_{x \to 0} e^{2 x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.4

Move the limit into the exponent.

$\frac{e^{4 lim_{x \to 0} x} - e^{lim_{x \to 0} 2 x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.5

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

$\frac{e^{4 lim_{x \to 0} x} - e^{2 lim_{x \to 0} x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.6

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

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Step 1.1.2.6.1

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

$\frac{e^{4 \cdot 0} - e^{2 lim_{x \to 0} x}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.6.2

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

$\frac{e^{4 \cdot 0} - e^{2 \cdot 0}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.7

Simplify the answer.

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Step 1.1.2.7.1

Simplify each term.

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Step 1.1.2.7.1.1

Multiply $4$ by $0$ .

$\frac{e^{0} - e^{2 \cdot 0}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.7.1.2

Anything raised to $0$ is $1$ .

$\frac{1 - e^{2 \cdot 0}}{lim_{x \to 0} e^{x} - 1}$

Step 1.1.2.7.1.3

Multiply $2$ by $0$ .

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

Step 1.1.2.7.1.4

Anything raised to $0$ is $1$ .

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

Step 1.1.2.7.1.5

Multiply $- 1$ by $1$ .

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

Step 1.1.2.7.2

Subtract $1$ from $1$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

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

Step 1.1.3.1.2

Move the limit into the exponent.

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

$\frac{0}{e^{0} - 1 \cdot 1}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Anything raised to $0$ is $1$ .

$\frac{0}{1 - 1 \cdot 1}$

Step 1.1.3.3.1.2

Multiply $- 1$ by $1$ .

$\frac{0}{1 - 1}$

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.1.4

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

Undefined

$\frac{0}{0}$

Step 1.2

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.

$lim_{x \to 0} \frac{e^{4 x} - e^{2 x}}{e^{x} - 1} = lim_{x \to 0} \frac{\frac{d}{d x} [e^{4 x} - e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{4 x} - e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.2

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

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [e^{4 x}]$ .

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Step 1.3.3.1

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

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Step 1.3.3.1.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$lim_{x \to 0} \frac{\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x] + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{e^{u_{1}} \frac{d}{d x} [4 x] + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.3.1.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$lim_{x \to 0} \frac{e^{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{e^{4 x} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.3.3

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

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

Step 1.3.3.4

Multiply $4$ by $1$ .

$lim_{x \to 0} \frac{e^{4 x} \cdot 4 + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.3.5

Move $4$ to the left of $e^{4 x}$ .

$lim_{x \to 0} \frac{4 e^{4 x} + \frac{d}{d x} [- e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- e^{2 x}]$ .

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Step 1.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{2 x}$ with respect to $x$ is $- \frac{d}{d x} [e^{2 x}]$ .

$lim_{x \to 0} \frac{4 e^{4 x} - \frac{d}{d x} [e^{2 x}]}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.2

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

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Step 1.3.4.2.1

To apply the Chain Rule, set $u_{2}$ as $2 x$ .

$lim_{x \to 0} \frac{4 e^{4 x} - (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [2 x])}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{4 e^{4 x} - (e^{u_{2}} \frac{d}{d x} [2 x])}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.2.3

Replace all occurrences of $u_{2}$ with $2 x$ .

$lim_{x \to 0} \frac{4 e^{4 x} - (e^{2 x} \frac{d}{d x} [2 x])}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.3

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

$lim_{x \to 0} \frac{4 e^{4 x} - (e^{2 x} (2 \frac{d}{d x} [x]))}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.4

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

$lim_{x \to 0} \frac{4 e^{4 x} - (e^{2 x} (2 \cdot 1))}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.5

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{4 e^{4 x} - (e^{2 x} \cdot 2)}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.6

Move $2$ to the left of $e^{2 x}$ .

$lim_{x \to 0} \frac{4 e^{4 x} - (2 \cdot e^{2 x})}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.4.7

Multiply $2$ by $- 1$ .

$lim_{x \to 0} \frac{4 e^{4 x} - 2 e^{2 x}}{\frac{d}{d x} [e^{x} - 1]}$

Step 1.3.5

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

$lim_{x \to 0} \frac{4 e^{4 x} - 2 e^{2 x}}{\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 1]}$

Step 1.3.6

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

$lim_{x \to 0} \frac{4 e^{4 x} - 2 e^{2 x}}{e^{x} + \frac{d}{d x} [- 1]}$

Step 1.3.7

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

$lim_{x \to 0} \frac{4 e^{4 x} - 2 e^{2 x}}{e^{x} + 0}$

Step 1.3.8

Add $e^{x}$ and $0$ .

$lim_{x \to 0} \frac{4 e^{4 x} - 2 e^{2 x}}{e^{x}}$

Step 2

Evaluate the limit.

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Step 2.1

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

$\frac{lim_{x \to 0} 4 e^{4 x} - 2 e^{2 x}}{lim_{x \to 0} e^{x}}$

Step 2.2

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

$\frac{lim_{x \to 0} 4 e^{4 x} - lim_{x \to 0} 2 e^{2 x}}{lim_{x \to 0} e^{x}}$

Step 2.3

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 lim_{x \to 0} e^{4 x} - lim_{x \to 0} 2 e^{2 x}}{lim_{x \to 0} e^{x}}$

Step 2.4

Move the limit into the exponent.

$\frac{4 e^{lim_{x \to 0} 4 x} - lim_{x \to 0} 2 e^{2 x}}{lim_{x \to 0} e^{x}}$

Step 2.5

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{4 e^{4 lim_{x \to 0} x} - lim_{x \to 0} 2 e^{2 x}}{lim_{x \to 0} e^{x}}$

Step 2.6

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

$\frac{4 e^{4 lim_{x \to 0} x} - 2 lim_{x \to 0} e^{2 x}}{lim_{x \to 0} e^{x}}$

Step 2.7

Move the limit into the exponent.

$\frac{4 e^{4 lim_{x \to 0} x} - 2 e^{lim_{x \to 0} 2 x}}{lim_{x \to 0} e^{x}}$

Step 2.8

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

$\frac{4 e^{4 lim_{x \to 0} x} - 2 e^{2 lim_{x \to 0} x}}{lim_{x \to 0} e^{x}}$

Step 2.9

Move the limit into the exponent.

$\frac{4 e^{4 lim_{x \to 0} x} - 2 e^{2 lim_{x \to 0} x}}{e^{lim_{x \to 0} x}}$

Step 3

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

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Step 3.1

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

$\frac{4 e^{4 \cdot 0} - 2 e^{2 lim_{x \to 0} x}}{e^{lim_{x \to 0} x}}$

Step 3.2

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

$\frac{4 e^{4 \cdot 0} - 2 e^{2 \cdot 0}}{e^{lim_{x \to 0} x}}$

Step 3.3

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

$\frac{4 e^{4 \cdot 0} - 2 e^{2 \cdot 0}}{e^{0}}$

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $4$ by $0$ .

$\frac{4 e^{0} - 2 e^{2 \cdot 0}}{e^{0}}$

Step 4.1.2

Anything raised to $0$ is $1$ .

$\frac{4 \cdot 1 - 2 e^{2 \cdot 0}}{e^{0}}$

Step 4.1.3

Multiply $4$ by $1$ .

$\frac{4 - 2 e^{2 \cdot 0}}{e^{0}}$

Step 4.1.4

Multiply $2$ by $0$ .

$\frac{4 - 2 e^{0}}{e^{0}}$

Step 4.1.5

Anything raised to $0$ is $1$ .

$\frac{4 - 2 \cdot 1}{e^{0}}$

Step 4.1.6

Multiply $- 2$ by $1$ .

$\frac{4 - 2}{e^{0}}$

Step 4.1.7

Subtract $2$ from $4$ .

$\frac{2}{e^{0}}$

Step 4.2

Anything raised to $0$ is $1$ .

$\frac{2}{1}$

Step 4.3

Divide $2$ by $1$ .

$2$