question\nevaluate the limit: $lim_{x\rightarrow - 2^{+}}\frac{sqrt{x + 2}}{x + 2}$

Answer:

$\infty$

Explanation:

Step1: Let $t=x + 2$

As $x\to - 2^{+}$, then $t\to0^{+}$. And $x=t - 2$. The limit becomes $\lim_{t\to0^{+}}\frac{\sqrt{t}}{t}$.

Step2: Simplify the expression

$\lim_{t\to0^{+}}\frac{\sqrt{t}}{t}=\lim_{t\to0^{+}}\frac{\sqrt{t}}{(\sqrt{t})^2}=\lim_{t\to0^{+}}\frac{1}{\sqrt{t}}$.

Step3: Evaluate the limit

As $t\to0^{+}$, $\frac{1}{\sqrt{t}}\to\infty$.