Calculus Examples

$lim_{x \to 0} \frac{{(2 + x)}^{3} - 8}{x}$

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.

$\frac{lim_{x \to 0} {(2 + x)}^{3} - 8}{lim_{x \to 0} x}$

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 $0$ .

$\frac{lim_{x \to 0} {(2 + x)}^{3} - lim_{x \to 0} 8}{lim_{x \to 0} x}$

Step 1.1.2.1.2

Move the exponent $3$ from ${(2 + x)}^{3}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 0} 2 + x)}^{3} - lim_{x \to 0} 8}{lim_{x \to 0} x}$

Step 1.1.2.1.3

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

$\frac{{(lim_{x \to 0} 2 + lim_{x \to 0} x)}^{3} - lim_{x \to 0} 8}{lim_{x \to 0} x}$

Step 1.1.2.1.4

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

$\frac{{(2 + lim_{x \to 0} x)}^{3} - lim_{x \to 0} 8}{lim_{x \to 0} x}$

Step 1.1.2.1.5

Evaluate the limit of $8$ which is constant as $x$ approaches $0$ .

$\frac{{(2 + lim_{x \to 0} x)}^{3} - 1 \cdot 8}{lim_{x \to 0} x}$

Step 1.1.2.2

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

$\frac{{(2 + 0)}^{3} - 1 \cdot 8}{lim_{x \to 0} x}$

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

Add $2$ and $0$ .

$\frac{2^{3} - 1 \cdot 8}{lim_{x \to 0} x}$

Step 1.1.2.3.1.2

Raise $2$ to the power of $3$ .

$\frac{8 - 1 \cdot 8}{lim_{x \to 0} x}$

Step 1.1.2.3.1.3

Multiply $- 1$ by $8$ .

$\frac{8 - 8}{lim_{x \to 0} x}$

Step 1.1.2.3.2

Subtract $8$ from $8$ .

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

Step 1.1.3

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

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{{(2 + x)}^{3} - 8}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [{(2 + x)}^{3} - 8]}{\frac{d}{d x} [x]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [{(2 + x)}^{3} - 8]}{\frac{d}{d x} [x]}$

Step 1.3.2

By the Sum Rule, the derivative of ${(2 + x)}^{3} - 8$ with respect to $x$ is $\frac{d}{d x} [{(2 + x)}^{3}] + \frac{d}{d x} [- 8]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [{(2 + x)}^{3}] + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [{(2 + x)}^{3}]$ .

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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 $f (x) = x^{3}$ and $g (x) = 2 + x$ .

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Step 1.3.3.1.1

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

$lim_{x \to 0} \frac{\frac{d}{d u} [u^{3}] \frac{d}{d x} [2 + x] + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$lim_{x \to 0} \frac{3 u^{2} \frac{d}{d x} [2 + x] + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3.1.3

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

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} \frac{d}{d x} [2 + x] + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3.2

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

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} (\frac{d}{d x} [2] + \frac{d}{d x} [x]) + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3.3

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

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} (0 + \frac{d}{d x} [x]) + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

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 $n = 1$ .

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} (0 + 1) + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3.5

Add $0$ and $1$ .

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} \cdot 1 + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.3.6

Multiply $3$ by $1$ .

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} + \frac{d}{d x} [- 8]}{\frac{d}{d x} [x]}$

Step 1.3.4

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$lim_{x \to 0} \frac{3 {(2 + x)}^{2} + 0}{\frac{d}{d x} [x]}$

Step 1.3.5

Add $3 {(2 + x)}^{2}$ and $0$ .

$lim_{x \to 0} \frac{3 {(2 + x)}^{2}}{\frac{d}{d x} [x]}$

Step 1.3.6

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

$lim_{x \to 0} \frac{3 {(2 + x)}^{2}}{1}$

Step 1.4

Divide $3 {(2 + x)}^{2}$ by $1$ .

$lim_{x \to 0} 3 {(2 + x)}^{2}$

Step 2

Evaluate the limit.

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Step 2.1

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$3 lim_{x \to 0} {(2 + x)}^{2}$

Step 2.2

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

$3 {(lim_{x \to 0} 2 + x)}^{2}$

Step 2.3

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

$3 {(lim_{x \to 0} 2 + lim_{x \to 0} x)}^{2}$

Step 2.4

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

$3 {(2 + lim_{x \to 0} x)}^{2}$

Step 3

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

$3 {(2 + 0)}^{2}$

Step 4

Simplify the answer.

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Step 4.1

Add $2$ and $0$ .

$3 \cdot 2^{2}$

Step 4.2

Raise $2$ to the power of $2$ .

$3 \cdot 4$

Step 4.3

Multiply $3$ by $4$ .

$12$