Calculus Examples

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

Step 1

Move the term $\frac{1}{- 2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

Evaluate the limit.

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

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

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

Step 2.1.2.1.3

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

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

Step 2.1.2.1.4

Evaluate the limit of $4$ which is constant as $x$ approaches $0$ .

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

Step 2.1.2.1.5

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Simplify the answer.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Multiply $- 1$ by $4$ .

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

Step 2.1.2.3.1.2

Add $- 4$ and $0$ .

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

Step 2.1.2.3.1.3

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

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

Step 2.1.2.3.1.4

Multiply $- 1$ by $16$ .

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

Step 2.1.2.3.2

Subtract $16$ from $16$ .

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

Step 2.1.3

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

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

Step 2.1.4

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

Undefined

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

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

Rewrite ${(- 4 + x)}^{2}$ as $(- 4 + x) (- 4 + x)$ .

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

Step 2.3.3

Expand $(- 4 + x) (- 4 + x)$ using the FOIL Method.

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Step 2.3.3.1

Apply the distributive property.

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

Step 2.3.3.2

Apply the distributive property.

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

Step 2.3.3.3

Apply the distributive property.

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

Step 2.3.4

Simplify and combine like terms.

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Step 2.3.4.1

Simplify each term.

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Step 2.3.4.1.1

Multiply $- 4$ by $- 4$ .

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

Step 2.3.4.1.2

Move $- 4$ to the left of $x$ .

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

Step 2.3.4.1.3

Multiply $x$ by $x$ .

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

Step 2.3.4.2

Subtract $4 x$ from $- 4 x$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{\frac{d}{d x} [16 - 8 x + x^{2} - 16]}{\frac{d}{d x} [x]}$

Step 2.3.5

By the Sum Rule, the derivative of $16 - 8 x + x^{2} - 16$ with respect to $x$ is $\frac{d}{d x} [16] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{\frac{d}{d x} [16] + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 2.3.6

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

$\frac{1}{- 2} lim_{x \to 0} \frac{0 + \frac{d}{d x} [- 8 x] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 2.3.7

Evaluate $\frac{d}{d x} [- 8 x]$ .

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Step 2.3.7.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x$ with respect to $x$ is $- 8 \frac{d}{d x} [x]$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{0 - 8 \frac{d}{d x} [x] + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 2.3.7.2

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

$\frac{1}{- 2} lim_{x \to 0} \frac{0 - 8 \cdot 1 + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 2.3.7.3

Multiply $- 8$ by $1$ .

$\frac{1}{- 2} lim_{x \to 0} \frac{0 - 8 + \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 2.3.8

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

$\frac{1}{- 2} lim_{x \to 0} \frac{0 - 8 + 2 x + \frac{d}{d x} [- 16]}{\frac{d}{d x} [x]}$

Step 2.3.9

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

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

Step 2.3.10

Simplify.

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Step 2.3.10.1

Combine terms.

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Step 2.3.10.1.1

Subtract $8$ from $0$ .

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

Step 2.3.10.1.2

Add $- 8 + 2 x$ and $0$ .

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

Step 2.3.10.2

Reorder terms.

$\frac{1}{- 2} lim_{x \to 0} \frac{2 x - 8}{\frac{d}{d x} [x]}$

Step 2.3.11

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

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

Step 2.4

Divide $2 x - 8$ by $1$ .

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

Step 3

Evaluate the limit.

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Step 3.1

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

$\frac{1}{- 2} (lim_{x \to 0} 2 x - lim_{x \to 0} 8)$

Step 3.2

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

$\frac{1}{- 2} (2 lim_{x \to 0} x - lim_{x \to 0} 8)$

Step 3.3

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

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

Step 4

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

$\frac{1}{- 2} (2 \cdot 0 - 1 \cdot 8)$

Step 5

Simplify the answer.

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Step 5.1

Move the negative in front of the fraction.

$- \frac{1}{2} (2 \cdot 0 - 1 \cdot 8)$

Step 5.2

Simplify each term.

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Step 5.2.1

Multiply $2$ by $0$ .

$- \frac{1}{2} (0 - 1 \cdot 8)$

Step 5.2.2

Multiply $- 1$ by $8$ .

$- \frac{1}{2} (0 - 8)$

Step 5.3

Subtract $8$ from $0$ .

$- \frac{1}{2} \cdot - 8$

Step 5.4

Cancel the common factor of $2$ .

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Step 5.4.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$\frac{- 1}{2} \cdot - 8$

Step 5.4.2

Factor $2$ out of $- 8$ .

$\frac{- 1}{2} \cdot (2 (- 4))$

Step 5.4.3

Cancel the common factor.

$\frac{- 1}{2} \cdot (2 \cdot - 4)$

Step 5.4.4

Rewrite the expression.

$- 1 \cdot - 4$

Step 5.5

Multiply $- 1$ by $- 4$ .

$4$