Write the Fraction in Simplest Form limit as x approaches 0 of (e^(2x)-1)/(e^x-1)

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

Step 1

Apply L\'Hospital\'s rule.

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

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

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

Evaluate the limit of the numerator.

Evaluate the limit.

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

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

Move the limit into the exponent.

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

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

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

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

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

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

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

Simplify the answer.

Simplify each term.

Multiply $2$ by $0$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $1$ .

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

Evaluate the limit of the denominator.

Evaluate the limit.

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

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

Move the limit into the exponent.

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

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

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

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

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

Simplify the answer.

Simplify each term.

Anything raised to $0$ is $1$ .

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

Multiply $- 1$ by $1$ .

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

Subtract $1$ from $1$ .

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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

Undefined

$\frac{0}{0}$

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.

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

Find the derivative of the numerator and denominator.

Differentiate the numerator and denominator.

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

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

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

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

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 $g (x) = 2 x$ .

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

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

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

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $e^{2 x}$ and $e^{x}$ .

Factor $e^{x}$ out of $2 e^{2 x}$ .

$lim_{x \to 0} \frac{e^{x} (2 e^{x})}{e^{x}}$

Cancel the common factors.

Multiply by $1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Step 2

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

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

Step 3

Move the limit into the exponent.

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

Step 4

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

$2 e^{0}$

Step 5

Anything raised to $0$ is $1$ .

$2 \cdot 1$

Step 6

Multiply $2$ by $1$ .

$2$

Write the Fraction in Simplest Form limit as x approaches 0 of (e^(2x)-1)/(e^x-1)

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