Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

Move the limit into the exponent.

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

Step 1.1.2.1.3

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

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

Step 1.1.2.1.4

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Multiply $2$ by $0$ .

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

Step 1.1.2.3.1.2

Anything raised to $0$ is $1$ .

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

Step 1.1.2.3.1.3

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to 0} x} = 2$

Step 1.1.2.3.2

Subtract $1$ from $1$ .

$\frac{0}{lim_{x \to 0} x} = 2$

Step 1.1.3

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

$\frac{0}{0} = 2$

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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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 $g (x) = 2 x$ .

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Step 1.3.3.1.1

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

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

Step 1.3.3.1.2

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

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

Step 1.3.3.1.3

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

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

Step 1.3.3.2

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

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

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 $n = 1$ .

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

Step 1.3.3.4

Multiply $2$ by $1$ .

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

Step 1.3.3.5

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

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

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

Step 1.4

Divide $2 e^{2 x}$ by $1$ .

$lim_{x \to 0} 2 e^{2 x} = 2$

Step 2

Evaluate the limit.

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Step 2.1

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

$2 lim_{x \to 0} e^{2 x} = 2$

Step 2.2

Move the limit into the exponent.

$2 e^{lim_{x \to 0} 2 x} = 2$

Step 2.3

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

$2 e^{2 lim_{x \to 0} x} = 2$

Step 3

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

$2 e^{2 \cdot 0} = 2$

Step 4

Simplify the answer.

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Step 4.1

Multiply $2$ by $0$ .

$2 e^{0} = 2$

Step 4.2

Anything raised to $0$ is $1$ .

$2 \cdot 1 = 2$

Step 4.3

Multiply $2$ by $1$ .

$2 = 2$