Calculus Examples

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

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.

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

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 $0$ .

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

Step 1.1.2.1.2

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

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

Step 1.1.2.1.3

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

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

Step 1.1.2.1.4

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

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

Step 1.1.2.1.5

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

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

Step 1.1.2.1.6

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

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

Step 1.1.2.2

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

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

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

Simplify each term.

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Step 1.1.2.3.1.1.1

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

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

Step 1.1.2.3.1.1.2

Multiply $- 1$ by $4$ .

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

Step 1.1.2.3.1.2

Subtract $4$ from $0$ .

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

Step 1.1.2.3.1.3

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

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

Step 1.1.2.3.1.4

Multiply $- 1$ by $16$ .

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

Step 1.1.2.3.2

Subtract $16$ from $16$ .

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

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

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

$\frac{0}{0}$

Step 1.1.3.4

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

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Step 1.3.3.1.1

To apply the Chain Rule, set $u$ as $x^{2} - 4$ .

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

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 $n = 2$ .

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

Step 1.3.3.1.3

Replace all occurrences of $u$ with $x^{2} - 4$ .

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

Step 1.3.3.2

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

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

Step 1.3.3.3

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

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

Step 1.3.3.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

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

Step 1.3.3.5

Add $2 x$ and $0$ .

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

Step 1.3.3.6

Multiply $2$ by $2$ .

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

Step 1.3.4

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

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

Step 1.3.5

Simplify.

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Step 1.3.5.1

Apply the distributive property.

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

Step 1.3.5.2

Apply the distributive property.

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

Step 1.3.5.3

Combine terms.

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Step 1.3.5.3.1

Raise $x$ to the power of $1$ .

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

Step 1.3.5.3.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.3.5.3.3

Add $1$ and $2$ .

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

Step 1.3.5.3.4

Multiply $4$ by $- 4$ .

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

Step 1.3.5.3.5

Add $4 x^{3} - 16 x$ and $0$ .

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

Step 1.3.6

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

$lim_{x \to 0} \frac{4 x^{3} - 16 x}{2 x}$

Step 2

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

$\frac{1}{2} lim_{x \to 0} \frac{4 x^{3} - 16 x}{x}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

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

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Step 3.1.1

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

$\frac{1}{2} \cdot \frac{lim_{x \to 0} 4 x^{3} - 16 x}{lim_{x \to 0} x}$

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

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

$\frac{1}{2} \cdot \frac{lim_{x \to 0} 4 x^{3} - lim_{x \to 0} 16 x}{lim_{x \to 0} x}$

Step 3.1.2.2

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

$\frac{1}{2} \cdot \frac{4 lim_{x \to 0} x^{3} - lim_{x \to 0} 16 x}{lim_{x \to 0} x}$

Step 3.1.2.3

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

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

Step 3.1.2.4

Move the term $16$ outside of the limit because it is constant with respect to $x$ .

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

Step 3.1.2.5

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

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Step 3.1.2.5.1

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

$\frac{1}{2} \cdot \frac{4 \cdot 0^{3} - 16 lim_{x \to 0} x}{lim_{x \to 0} x}$

Step 3.1.2.5.2

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

$\frac{1}{2} \cdot \frac{4 \cdot 0^{3} - 16 \cdot 0}{lim_{x \to 0} x}$

Step 3.1.2.6

Simplify the answer.

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Step 3.1.2.6.1

Simplify each term.

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Step 3.1.2.6.1.1

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

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

Step 3.1.2.6.1.2

Multiply $4$ by $0$ .

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

Step 3.1.2.6.1.3

Multiply $- 16$ by $0$ .

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

Step 3.1.2.6.2

Add $0$ and $0$ .

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

Step 3.1.3

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

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

Step 3.1.4

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

Undefined

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

Step 3.2

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.

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

Step 3.3.2

By the Sum Rule, the derivative of $4 x^{3} - 16 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 16 x]$ .

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

Step 3.3.3

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 3.3.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

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

Step 3.3.3.2

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

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

Step 3.3.3.3

Multiply $3$ by $4$ .

$\frac{1}{2} lim_{x \to 0} \frac{12 x^{2} + \frac{d}{d x} [- 16 x]}{\frac{d}{d x} [x]}$

Step 3.3.4

Evaluate $\frac{d}{d x} [- 16 x]$ .

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Step 3.3.4.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16 x$ with respect to $x$ is $- 16 \frac{d}{d x} [x]$ .

$\frac{1}{2} lim_{x \to 0} \frac{12 x^{2} - 16 \frac{d}{d x} [x]}{\frac{d}{d x} [x]}$

Step 3.3.4.2

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

$\frac{1}{2} lim_{x \to 0} \frac{12 x^{2} - 16 \cdot 1}{\frac{d}{d x} [x]}$

Step 3.3.4.3

Multiply $- 16$ by $1$ .

$\frac{1}{2} lim_{x \to 0} \frac{12 x^{2} - 16}{\frac{d}{d x} [x]}$

Step 3.3.5

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

$\frac{1}{2} lim_{x \to 0} \frac{12 x^{2} - 16}{1}$

Step 3.4

Divide $12 x^{2} - 16$ by $1$ .

$\frac{1}{2} lim_{x \to 0} 12 x^{2} - 16$

Step 4

Evaluate the limit.

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Step 4.1

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

$\frac{1}{2} (lim_{x \to 0} 12 x^{2} - lim_{x \to 0} 16)$

Step 4.2

Move the term $12$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2} (12 lim_{x \to 0} x^{2} - lim_{x \to 0} 16)$

Step 4.3

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

$\frac{1}{2} (12 {(lim_{x \to 0} x)}^{2} - lim_{x \to 0} 16)$

Step 4.4

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

$\frac{1}{2} (12 {(lim_{x \to 0} x)}^{2} - 1 \cdot 16)$

Step 5

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

$\frac{1}{2} (12 \cdot 0^{2} - 1 \cdot 16)$

Step 6

Simplify the answer.

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Step 6.1

Simplify each term.

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Step 6.1.1

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

$\frac{1}{2} (12 \cdot 0 - 1 \cdot 16)$

Step 6.1.2

Multiply $12$ by $0$ .

$\frac{1}{2} (0 - 1 \cdot 16)$

Step 6.1.3

Multiply $- 1$ by $16$ .

$\frac{1}{2} (0 - 16)$

Step 6.2

Subtract $16$ from $0$ .

$\frac{1}{2} \cdot - 16$

Step 6.3

Cancel the common factor of $2$ .

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Step 6.3.1

Factor $2$ out of $- 16$ .

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

Step 6.3.2

Cancel the common factor.

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

Step 6.3.3

Rewrite the expression.

$- 8$