Find $\lim\limits_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$.

Question

Find $\displaystyle \lim_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$.

I tried to rationalize it, but doesn't help either. Please give me some hints. Thank you.

Answer 1

Hint: $$ \frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}=\left(\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}\frac {\sqrt{x+3}+2}{\sqrt{x+8}+3}\right)\frac {\sqrt{x+8}+3}{\sqrt{x+3}+2} $$

Answer 2

I suggest using heavy artillery. Denote $y=x-1$ for convenience. We need to find $$ \lim_{y \to 0}\frac{\sqrt{4+y}-2}{\sqrt{9+y}-3}. $$ Not you just need to use the fact $$ \sqrt{a^2+y} = a + \frac{y}{2a} + o(y) $$ when $y \to 0$.