Calculus Examples

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

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.

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

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 $0$ .

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

Move the limit into the exponent.

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

Step 1.1.2.1.4

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

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

Step 1.1.2.1.5

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

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

Step 1.1.2.2

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

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

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 $0$ .

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

Step 1.1.2.3.1.2

Anything raised to $0$ is $1$ .

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

Step 1.1.2.3.1.3

Multiply $2$ by $1$ .

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

Step 1.1.2.3.1.4

Multiply $- 1$ by $2$ .

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

Step 1.1.2.3.2

Subtract $2$ from $2$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

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

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

Step 1.1.3.2

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

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

Step 1.1.3.3

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

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Step 1.1.3.3.1

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

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

Step 1.1.3.3.2

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

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

Step 1.1.3.4

Simplify the answer.

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Step 1.1.3.4.1

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0 + 0}$

Step 1.1.3.4.2

Add $0$ and $0$ .

$\frac{0}{0}$

Step 1.1.3.4.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.5

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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Step 1.3.3.1

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

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

Step 1.3.3.2

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 1.3.3.2.1

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

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

Step 1.3.3.2.2

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

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

Step 1.3.3.2.3

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

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

Step 1.3.3.3

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

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

Step 1.3.3.4

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

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

Step 1.3.3.5

Multiply $2$ by $1$ .

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

Step 1.3.3.6

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

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

Step 1.3.3.7

Multiply $2$ by $2$ .

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

Step 1.3.4

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 1.3.5

Add $4 e^{2 x}$ and $0$ .

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

Step 1.3.6

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

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

Step 1.3.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 1.3.8

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

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

Step 2

Evaluate the limit.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Move the limit into the exponent.

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

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

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

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 $x$ .

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

Step 3.2

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

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

Step 4

Simplify the answer.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Multiply $2$ by $0$ .

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

Step 4.1.2

Anything raised to $0$ is $1$ .

$4 \frac{1}{2 \cdot 0 + 1}$

Step 4.2

Simplify the denominator.

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Step 4.2.1

Multiply $2$ by $0$ .

$4 \frac{1}{0 + 1}$

Step 4.2.2

Add $0$ and $1$ .

$4 (\frac{1}{1})$

Step 4.3

Cancel the common factor of $1$ .

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Step 4.3.1

Cancel the common factor.

$4 (\frac{1}{1})$

Step 4.3.2

Rewrite the expression.

$4 \cdot 1$

Step 4.4

Multiply $4$ by $1$ .

$4$