Question

question
evaluate the limit: $$\lim_{x \to -2} \frac{\sqrt{x + 3} - 1}{x + 2}$$

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x + 3}+1$.

$$\lim_{x ightarrow - 2}\frac{\sqrt{x + 3}-1}{x + 2}\times\frac{\sqrt{x + 3}+1}{\sqrt{x + 3}+1}$$

Using the formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(\sqrt{x + 3})^{2}-1^{2}=x + 3-1=x + 2$.
So the limit is $\lim_{x
ightarrow - 2}\frac{x + 2}{(x + 2)(\sqrt{x + 3}+1)}$.

Step2: Simplify the expression

Cancel out the common factor $(x + 2)$ (since $x
eq - 2$ as we are taking the limit as $x
ightarrow - 2$, not evaluating at $x=-2$).
The expression simplifies to $\lim_{x
ightarrow - 2}\frac{1}{\sqrt{x + 3}+1}$.

Step3: Substitute $x=-2$

Substitute $x=-2$ into $\frac{1}{\sqrt{x + 3}+1}$.
We get $\frac{1}{\sqrt{-2 + 3}+1}=\frac{1}{1 + 1}$.

Answer:

$\frac{1}{2}$