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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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    $\begingroup$ How about multiplying by the conjugate. $\endgroup$
    – Amzoti
    Mar 3, 2013 at 4:44
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    $\begingroup$ How did you rationalize it? $\endgroup$ Mar 3, 2013 at 4:44
  • $\begingroup$ @jasoncube you could see Calvin's answer here: math.stackexchange.com/questions/275990/… $\endgroup$ Mar 3, 2013 at 4:51
  • $\begingroup$ 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. $\endgroup$ Mar 3, 2013 at 5:08

2 Answers 2

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

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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 $y \to 0$.

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  • $\begingroup$ I have never used this approach when rationalising radical limits. $(+1)$ :D $\endgroup$
    – Mr Pie
    Oct 3, 2018 at 22:32

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