Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

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

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

Step 1.2.1.2

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

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

Step 1.2.2

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

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

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

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

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

Step 1.2.3.2

Multiply $4$ by $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

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

Step 1.3.2

Move the limit into the exponent.

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

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

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Step 1.3.6.1

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

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

Step 1.3.6.2

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

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

Step 1.3.7

Simplify the answer.

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Step 1.3.7.1

Simplify each term.

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Step 1.3.7.1.1

Multiply $4$ by $0$ .

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

Step 1.3.7.1.2

Anything raised to $0$ is $1$ .

$\frac{0}{1 - 4 \cdot 0 - 1 \cdot 1}$

Step 1.3.7.1.3

Multiply $- 4$ by $0$ .

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

Step 1.3.7.1.4

Multiply $- 1$ by $1$ .

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

Step 1.3.7.2

Add $1$ and $0$ .

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

Step 1.3.7.3

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.3.7.4

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

Undefined

$\frac{0}{0}$

Step 1.3.8

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

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

Step 3.3

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

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

Step 3.4

Multiply $2$ by $4$ .

$lim_{x \to 0} \frac{8 x}{\frac{d}{d x} [e^{4 x} - 4 x - 1]}$

Step 3.5

By the Sum Rule, the derivative of $e^{4 x} - 4 x - 1$ with respect to $x$ is $\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]$ .

$lim_{x \to 0} \frac{8 x}{\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6

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

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Step 3.6.1

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 3.6.1.1

To apply the Chain Rule, set $u$ as $4 x$ .

$lim_{x \to 0} \frac{8 x}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [4 x] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6.1.2

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

$lim_{x \to 0} \frac{8 x}{e^{u} \frac{d}{d x} [4 x] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6.1.3

Replace all occurrences of $u$ with $4 x$ .

$lim_{x \to 0} \frac{8 x}{e^{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6.2

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

$lim_{x \to 0} \frac{8 x}{e^{4 x} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6.3

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

$lim_{x \to 0} \frac{8 x}{e^{4 x} (4 \cdot 1) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6.4

Multiply $4$ by $1$ .

$lim_{x \to 0} \frac{8 x}{e^{4 x} \cdot 4 + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.6.5

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

$lim_{x \to 0} \frac{8 x}{4 e^{4 x} + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 1]}$

Step 3.7

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

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Step 3.7.1

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

$lim_{x \to 0} \frac{8 x}{4 e^{4 x} - 4 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]}$

Step 3.7.2

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

$lim_{x \to 0} \frac{8 x}{4 e^{4 x} - 4 \cdot 1 + \frac{d}{d x} [- 1]}$

Step 3.7.3

Multiply $- 4$ by $1$ .

$lim_{x \to 0} \frac{8 x}{4 e^{4 x} - 4 + \frac{d}{d x} [- 1]}$

Step 3.8

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

$lim_{x \to 0} \frac{8 x}{4 e^{4 x} - 4 + 0}$

Step 3.9

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

$lim_{x \to 0} \frac{8 x}{4 e^{4 x} - 4}$

Step 4

Evaluate the limit.

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Step 4.1

Cancel the common factor of $8$ and $4 e^{4 x} - 4$ .

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Step 4.1.1

Factor $4$ out of $8 x$ .

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

Step 4.1.2

Cancel the common factors.

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Step 4.1.2.1

Factor $4$ out of $4 e^{4 x}$ .

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

Step 4.1.2.2

Factor $4$ out of $- 4$ .

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

Step 4.1.2.3

Factor $4$ out of $4 (e^{4 x}) + 4 (- 1)$ .

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

Step 4.1.2.4

Cancel the common factor.

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

Step 4.1.2.5

Rewrite the expression.

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

Step 4.2

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

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

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

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

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

Step 5.1.3

Evaluate the limit of the denominator.

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Step 5.1.3.1

Evaluate the limit.

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Step 5.1.3.1.1

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

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

Step 5.1.3.1.2

Move the limit into the exponent.

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

Step 5.1.3.1.3

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

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

Step 5.1.3.1.4

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

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

Step 5.1.3.2

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

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

Step 5.1.3.3

Simplify the answer.

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Step 5.1.3.3.1

Simplify each term.

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Step 5.1.3.3.1.1

Multiply $4$ by $0$ .

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

Step 5.1.3.3.1.2

Anything raised to $0$ is $1$ .

$2 \frac{0}{1 - 1 \cdot 1}$

Step 5.1.3.3.1.3

Multiply $- 1$ by $1$ .

$2 \frac{0}{1 - 1}$

Step 5.1.3.3.2

Subtract $1$ from $1$ .

$2 (\frac{0}{0})$

Step 5.1.3.3.3

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

Undefined

$2 (\frac{0}{0})$

Step 5.1.3.4

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

Undefined

$2 (\frac{0}{0})$

Step 5.1.4

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

Undefined

$2 (\frac{0}{0})$

Step 5.2

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.

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

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

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

Step 5.3.3

By the Sum Rule, the derivative of $e^{4 x} - 1$ with respect to $x$ is $\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [- 1]$ .

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

Step 5.3.4

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

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Step 5.3.4.1

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 5.3.4.1.1

To apply the Chain Rule, set $u$ as $4 x$ .

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

Step 5.3.4.1.2

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

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

Step 5.3.4.1.3

Replace all occurrences of $u$ with $4 x$ .

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

Step 5.3.4.2

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

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

Step 5.3.4.3

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

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

Step 5.3.4.4

Multiply $4$ by $1$ .

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

Step 5.3.4.5

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

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

Step 5.3.5

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

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

Step 5.3.6

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

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

Step 6

Evaluate the limit.

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Step 6.1

Move the term $\frac{1}{4}$ outside of the limit because it is constant with respect to $x$ .

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

Move the limit into the exponent.

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

Step 6.5

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

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

Step 7

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

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

Step 8

Simplify the answer.

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Step 8.1

Cancel the common factor of $2$ .

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Step 8.1.1

Factor $2$ out of $4$ .

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

Step 8.1.2

Cancel the common factor.

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

Step 8.1.3

Rewrite the expression.

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

Step 8.2

Combine.

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

Step 8.3

Multiply $1$ by $1$ .

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

Step 8.4

Simplify the denominator.

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Step 8.4.1

Multiply $4$ by $0$ .

$\frac{1}{2 e^{0}}$

Step 8.4.2

Anything raised to $0$ is $1$ .

$\frac{1}{2 \cdot 1}$

Step 8.5

Multiply $2$ by $1$ .

$\frac{1}{2}$