question\nevaluate the limit: $$\\lim_{x \\to -2} \\frac{\\sqrt{x + 3} - 1}{x + 2}$$
question\nevaluate the limit: $$\\lim_{x \\to -2} \\frac{\\sqrt{x + 3} - 1}{x + 2}$$
Answer
Explanation:
Step1: Rationalize the numerator
Multiply the numerator and denominator by $\sqrt{x + 3}+1$. $$\lim_{x\rightarrow - 2}\frac{\sqrt{x + 3}-1}{x + 2}\times\frac{\sqrt{x + 3}+1}{\sqrt{x + 3}+1}$$ 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 $\lim_{x\rightarrow - 2}\frac{x + 2}{(x + 2)(\sqrt{x + 3}+1)}$.
Step2: Simplify the expression
Cancel out the common factor $(x + 2)$ (since $x\neq - 2$ as we are taking the limit as $x\rightarrow - 2$, not evaluating at $x=-2$). The expression simplifies to $\lim_{x\rightarrow - 2}\frac{1}{\sqrt{x + 3}+1}$.
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:
$\frac{1}{2}$