Precalculus Examples

$lim_{x \to 0} \frac{{(2 - x)}^{4} - 16}{x}$

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.

$\frac{lim_{x \to 0} {(2 - x)}^{4} - 16}{lim_{x \to 0} x}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

$\frac{lim_{x \to 0} {(2 - x)}^{4} - lim_{x \to 0} 16}{lim_{x \to 0} x}$

Step 1.1.2.2

Move the exponent $4$ from ${(2 - x)}^{4}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 0} 2 - x)}^{4} - lim_{x \to 0} 16}{lim_{x \to 0} x}$

Step 1.1.2.3

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

$\frac{{(lim_{x \to 0} 2 - lim_{x \to 0} x)}^{4} - lim_{x \to 0} 16}{lim_{x \to 0} x}$

Step 1.1.2.4

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

$\frac{{(2 - lim_{x \to 0} x)}^{4} - lim_{x \to 0} 16}{lim_{x \to 0} x}$

Step 1.1.2.5

Evaluate the limit of $16$ which is constant as $x$ approaches $0$ .

$\frac{{(2 - lim_{x \to 0} x)}^{4} - 1 \cdot 16}{lim_{x \to 0} x}$

Step 1.1.2.6

Simplify terms.

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Step 1.1.2.6.1

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

$\frac{{(2 + 0)}^{4} - 1 \cdot 16}{lim_{x \to 0} x}$

Step 1.1.2.6.2

Simplify the answer.

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Step 1.1.2.6.2.1

Simplify each term.

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Step 1.1.2.6.2.1.1

Add $2$ and $0$ .

$\frac{2^{4} - 1 \cdot 16}{lim_{x \to 0} x}$

Step 1.1.2.6.2.1.2

Raise $2$ to the power of $4$ .

$\frac{16 - 1 \cdot 16}{lim_{x \to 0} x}$

Step 1.1.2.6.2.1.3

Multiply $- 1$ by $16$ .

$\frac{16 - 16}{lim_{x \to 0} x}$

Step 1.1.2.6.2.2

Subtract $16$ from $16$ .

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

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.

$lim_{x \to 0} \frac{{(2 - x)}^{4} - 16}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [{(2 - x)}^{4} - 16]}{\frac{d}{d x} [x]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [{(2 - x)}^{4} - 16]}{\frac{d}{d x} [x]}$

Step 1.3.2

By the Sum Rule, the derivative of ${(2 - x)}^{4} - 16$ with respect to $x$ is $\frac{d}{d x} [{(2 - x)}^{4}] + \frac{d}{d x} [- 16]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [{(2 - x)}^{4}] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3

Evaluate $\frac{d}{d x} [{(2 - x)}^{4}]$ .

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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 $f (x) = x^{4}$ and $g (x) = 2 - x$ .

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Step 1.3.3.1.1

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

$lim_{x \to 0} \frac{\frac{d}{d u} [u^{4}] \frac{d}{d x} [2 - x] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

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 $n = 4$ .

$lim_{x \to 0} \frac{4 u^{3} \frac{d}{d x} [2 - x] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.1.3

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

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} \frac{d}{d x} [2 - x] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.2

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

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} (\frac{d}{d x} [2] + \frac{d}{d x} [- x]) + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.3

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

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} (0 + \frac{d}{d x} [- x]) + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} (0 - \frac{d}{d x} [x]) + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.5

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

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} (0 - 1 \cdot 1) + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.6

Multiply $- 1$ by $1$ .

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} (0 - 1) + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.7

Subtract $1$ from $0$ .

$lim_{x \to 0} \frac{4 {(2 - x)}^{3} \cdot - 1 + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.3.8

Multiply $- 1$ by $4$ .

$lim_{x \to 0} \frac{- 4 {(2 - x)}^{3} + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 1.3.4

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$lim_{x \to 0} \frac{- 4 {(2 - x)}^{3} + 0}{\frac{d}{d x} [x]}$

Step 1.3.5

Add $- 4 {(2 - x)}^{3}$ and $0$ .

$lim_{x \to 0} \frac{- 4 {(2 - x)}^{3}}{\frac{d}{d x} [x]}$

Step 1.3.6

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

$lim_{x \to 0} \frac{- 4 {(2 - x)}^{3}}{1}$

Step 1.4

Divide $- 4 {(2 - x)}^{3}$ by $1$ .

$lim_{x \to 0} - 4 {(2 - x)}^{3}$

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 $x$ .

$- 4 lim_{x \to 0} {(2 - x)}^{3}$

Step 2.2

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

$- 4 {(lim_{x \to 0} 2 - x)}^{3}$

Step 2.3

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

$- 4 {(lim_{x \to 0} 2 - lim_{x \to 0} x)}^{3}$

Step 2.4

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

$- 4 {(2 - lim_{x \to 0} x)}^{3}$

Step 2.5

Simplify terms.

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Step 2.5.1

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

$- 4 {(2 + 0)}^{3}$

Step 2.5.2

Simplify the answer.

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Step 2.5.2.1

Add $2$ and $0$ .

$- 4 \cdot 2^{3}$

Step 2.5.2.2

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

$- 4 \cdot 8$

Step 2.5.2.3

Multiply $- 4$ by $8$ .

$- 32$