Question

question
evaluate the limit: $$\lim_{x\to4}\frac{-8 + 2x}{2 - \sqrt{x}}$$

Explanation:

Step1: Factor numerator and rationalize denominator

Factor numerator: $-8 + 2x=2(x - 4)$.
Multiply numerator and denominator by $2+\sqrt{x}$:

$$\lim_{x ightarrow4}\frac{2(x - 4)(2+\sqrt{x})}{(2-\sqrt{x})(2+\sqrt{x})}$$

Step2: Simplify using difference - of - squares formula

By $(a - b)(a + b)=a^{2}-b^{2}$, $(2-\sqrt{x})(2+\sqrt{x})=4 - x=-(x - 4)$.
So the limit becomes:

$$\lim_{x ightarrow4}\frac{2(x - 4)(2+\sqrt{x})}{-(x - 4)}$$

Cancel out $(x - 4)$ (since $x
eq4$ as $x
ightarrow4$):

$$\lim_{x ightarrow4}-2(2+\sqrt{x})$$

Step3: Substitute $x = 4$

Substitute $x = 4$ into $-2(2+\sqrt{x})$:
$-2(2+\sqrt{4})=-2(2 + 2)=-8$

Answer:

$-8$