Answer
(a)$$\lim_{x \to -2}\frac{x^2+x-2}{x+2}=-3$$
(b)$$\lim_{x \to 0}\frac{\sqrt{x+4}-2}{x}=\frac{1}{4}=0.25$$
Work Step by Step
(a)
Since by direct substituition we get the indeterminate form $\frac{0}{0}$, we should divide out the common factor as follows.$$\frac{x^2+x-2}{x+2}= \frac{(x+2)(x-1)}{x+2}= x-1 , \quad x \neq -2$$So, we obtain$$\lim_{x \to -2}\frac{x^2+x-2}{x+2}=\lim_{x \to -2}x-1= -3.$$
(b)
Since by direct substituition we get the indeterminate form $\frac{0}{0}$, we rationalize the numerator as follows.$$\frac{\sqrt{x+4}-2}{x}=\left ( \frac{\sqrt{x+4}-2}{x} \right ) \left ( \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} \right )=\frac{x+4-4}{x(\sqrt{x+4}+2)}=\frac{1}{\sqrt{x+4}+2}$$So, we obtain$$\lim_{x \to 0}\frac{\sqrt{x+4}-2}{x}= \lim_{x \to 0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{4}=0.25 \,.$$
