Calculus Examples

$lim_{x \to 4} \frac{x^{2} - 16}{2 - \sqrt{6}}$

Step 1

Evaluate the limit.

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Step 1.1

Move the term $\frac{1}{2 - \sqrt{6}}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{2 - \sqrt{6}} lim_{x \to 4} x^{2} - 16$

Step 1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $4$ .

$\frac{1}{2 - \sqrt{6}} (lim_{x \to 4} x^{2} - lim_{x \to 4} 16)$

Step 1.3

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

$\frac{1}{2 - \sqrt{6}} ({(lim_{x \to 4} x)}^{2} - lim_{x \to 4} 16)$

Step 1.4

Evaluate the limit of $16$ which is constant as $x$ approaches $4$ .

$\frac{1}{2 - \sqrt{6}} ({(lim_{x \to 4} x)}^{2} - 1 \cdot 16)$

Step 2

Evaluate the limit of $x$ by plugging in $4$ for $x$ .

$\frac{1}{2 - \sqrt{6}} (4^{2} - 1 \cdot 16)$

Step 3

Simplify the answer.

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Step 3.1

Multiply $\frac{1}{2 - \sqrt{6}}$ by $\frac{2 + \sqrt{6}}{2 + \sqrt{6}}$ .

$\frac{1}{2 - \sqrt{6}} \cdot \frac{2 + \sqrt{6}}{2 + \sqrt{6}} (4^{2} - 1 \cdot 16)$

Step 3.2

Multiply $\frac{1}{2 - \sqrt{6}}$ by $\frac{2 + \sqrt{6}}{2 + \sqrt{6}}$ .

$\frac{2 + \sqrt{6}}{(2 - \sqrt{6}) (2 + \sqrt{6})} (4^{2} - 1 \cdot 16)$

Step 3.3

Expand the denominator using the FOIL method.

$\frac{2 + \sqrt{6}}{4 + 2 \sqrt{6} - 2 \sqrt{6} - {\sqrt{6}}^{2}} (4^{2} - 1 \cdot 16)$

Step 3.4

Simplify.

$\frac{2 + \sqrt{6}}{- 2} (4^{2} - 1 \cdot 16)$

Step 3.5

Move the negative in front of the fraction.

$- \frac{2 + \sqrt{6}}{2} (4^{2} - 1 \cdot 16)$

Step 3.6

Simplify each term.

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Step 3.6.1

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

$- \frac{2 + \sqrt{6}}{2} (16 - 1 \cdot 16)$

Step 3.6.2

Multiply $- 1$ by $16$ .

$- \frac{2 + \sqrt{6}}{2} (16 - 16)$

Step 3.7

Subtract $16$ from $16$ .

$- \frac{2 + \sqrt{6}}{2} \cdot 0$

Step 3.8

Multiply $- \frac{2 + \sqrt{6}}{2} \cdot 0$ .

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Step 3.8.1

Multiply $0$ by $- 1$ .

$0 \frac{2 + \sqrt{6}}{2}$

Step 3.8.2

Multiply $0$ by $\frac{2 + \sqrt{6}}{2}$ .

$0$