Question

102.

$$ lim _ { x ightarrow - 3 } \frac { sqrt { x + 4 } - 1 } { x + 3 } $$

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x + 4}+1$.

$$\lim_{x ightarrow - 3}\frac{\sqrt{x + 4}-1}{x + 3}=\lim_{x ightarrow - 3}\frac{(\sqrt{x + 4}-1)(\sqrt{x + 4}+1)}{(x + 3)(\sqrt{x + 4}+1)}$$

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

Step2: Simplify the expression

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

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:

$\frac{1}{2}$