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.
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Martin Sleziak
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JSCB
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1How about multiplying by the conjugate. – Amzoti Mar 03 '13 at 04:44
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1How did you rationalize it? – lab bhattacharjee Mar 03 '13 at 04:44
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@jasoncube you could see Calvin's answer here: http://math.stackexchange.com/questions/275990/calculate-undersetx-rightarrow7-lim-frac-sqrtx2-sqrt3x20-sqrt/276012#276012 – Vincent Tjeng Mar 03 '13 at 04:51
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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. – Thomas Andrews Mar 03 '13 at 05:08
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Hint: $$ \frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}=\left(\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}\frac {\sqrt{x+3}+2}{\sqrt{x+8}+3}\right)\frac {\sqrt{x+8}+3}{\sqrt{x+3}+2} $$
Norbert
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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 $y \to 0$.
Dan Shved
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I have never used this approach when rationalising radical limits. $(+1)$ :D – Mr Pie Oct 03 '18 at 22:32