question\nevaluate 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 $2+\sqrt{x}$: $$\lim_{x\rightarrow4}\frac{2(x - 4)(2+\sqrt{x})}{(2-\sqrt{x})(2+\sqrt{x})}$$

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: $$\lim_{x\rightarrow4}\frac{2(x - 4)(2+\sqrt{x})}{-(x - 4)}$$ Cancel out $(x - 4)$ (since $x\neq4$ as $x\rightarrow4$): $$\lim_{x\rightarrow4}-2(2+\sqrt{x})$$

Step3: Substitute $x = 4$

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

Answer:

$-8$