102. $$ lim _ { x \rightarrow - 3 } \frac { sqrt { x + 4 } - 1 } { x + 3 } $$

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x + 4}+1$. $$\lim_{x\rightarrow - 3}\frac{\sqrt{x + 4}-1}{x + 3}=\lim_{x\rightarrow - 3}\frac{(\sqrt{x + 4}-1)(\sqrt{x + 4}+1)}{(x + 3)(\sqrt{x + 4}+1)}$$ Using the formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(x + 4)-1=x + 3$. So the limit is $\lim_{x\rightarrow - 3}\frac{x + 3}{(x + 3)(\sqrt{x + 4}+1)}$.

Step2: Simplify the expression

Cancel out the common factor $(x + 3)$ (since $x\neq - 3$ when taking the limit, $x\rightarrow - 3$ but $x\neq - 3$). We get $\lim_{x\rightarrow - 3}\frac{1}{\sqrt{x + 4}+1}$.

Step3: Substitute $x=-3$

Substitute $x = - 3$ into $\frac{1}{\sqrt{x + 4}+1}$. When $x=-3$, $\sqrt{-3 + 4}+1=\sqrt{1}+1=2$.

Answer:

$\frac{1}{2}$