Calculus Examples

$\frac{lim_{x \to 2} x^{3} - 8}{x^{4} - 16}$

Step 1

Evaluate the limit.

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Step 1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{lim_{x \to 2} x^{3} - lim_{x \to 2} 8}{x^{4} - 16}$

Step 1.2

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

$\frac{{(lim_{x \to 2} x)}^{3} - lim_{x \to 2} 8}{x^{4} - 16}$

Step 1.3

Evaluate the limit of $8$ which is constant as $x$ approaches $2$ .

$\frac{{(lim_{x \to 2} x)}^{3} - 1 \cdot 8}{x^{4} - 16}$

Step 2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{2^{3} - 1 \cdot 8}{x^{4} - 16}$

Step 3

Simplify the answer.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

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

$\frac{8 - 1 \cdot 8}{x^{4} - 16}$

Step 3.1.2

Multiply $- 1$ by $8$ .

$\frac{8 - 8}{x^{4} - 16}$

Step 3.1.3

Subtract $8$ from $8$ .

$\frac{0}{x^{4} - 16}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

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

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

Step 3.2.2

Rewrite $16$ as $4^{2}$ .

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

Step 3.2.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 4$ .

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

Step 3.2.4

Simplify.

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Step 3.2.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 3.2.4.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

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

Step 3.3

Divide $0$ by $(x^{2} + 4) (x + 2) (x - 2)$ .

$0$