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

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.

• How about multiplying by the conjugate. Mar 3, 2013 at 4:44
• How did you rationalize it? Mar 3, 2013 at 4:44
• @jasoncube you could see Calvin's answer here: math.stackexchange.com/questions/275990/… Mar 3, 2013 at 4:51
• Divide the numerator and denominator each by $x$ and you get a limit with a numerator and denominator that are derivatives of two different functions. Mar 3, 2013 at 5:08

Hint: $$\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}=\left(\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}\frac {\sqrt{x+3}+2}{\sqrt{x+8}+3}\right)\frac {\sqrt{x+8}+3}{\sqrt{x+3}+2}$$
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 $y \to 0$.
• I have never used this approach when rationalising radical limits. $(+1)$ :D Oct 3, 2018 at 22:32