iv) $lim_{x\rightarrow2}\frac{2 - sqrt{6 - x}}{sqrt{x + 2}-2}$
iv) $lim_{x\rightarrow2}\frac{2 - sqrt{6 - x}}{sqrt{x + 2}-2}$
Answer
Explanation:
Step1: Rationalize the numerator and denominator
Multiply the fraction by $\frac{(2 + \sqrt{6 - x})(\sqrt{x + 2}+2)}{(2 + \sqrt{6 - x})(\sqrt{x + 2}+2)}$. The original limit $\lim_{x\rightarrow2}\frac{2-\sqrt{6 - x}}{\sqrt{x + 2}-2}$ becomes $\lim_{x\rightarrow2}\frac{(2-\sqrt{6 - x})(2+\sqrt{6 - x})(\sqrt{x + 2}+2)}{(\sqrt{x + 2}-2)(\sqrt{x + 2}+2)(2+\sqrt{6 - x})}$. Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator is $(4-(6 - x))(\sqrt{x + 2}+2)=(x - 2)(\sqrt{x + 2}+2)$ and the denominator is $((x + 2)-4)(2+\sqrt{6 - x})=(x - 2)(2+\sqrt{6 - x})$.
Step2: Simplify the fraction
Cancel out the common factor $(x - 2)$ in the numerator and denominator. We get $\lim_{x\rightarrow2}\frac{\sqrt{x + 2}+2}{2+\sqrt{6 - x}}$.
Step3: Substitute $x = 2$
Substitute $x = 2$ into $\frac{\sqrt{x + 2}+2}{2+\sqrt{6 - x}}$. We have $\frac{\sqrt{2+2}+2}{2+\sqrt{6 - 2}}=\frac{2 + 2}{2+2}=1$.
Answer:
$1$