Answers

2015-04-12T17:23:07+05:30
\lim\limits_{x\to 1}\dfrac{\sqrt{ax+b}-\sqrt{x+1}}{x^2-1}=4\sqrt{2}

For the limit to be finite the numerator has to be 0 : 

\sqrt{a(1)+b}-\sqrt{1+1} = 2 \implies a+b = 2 \implies b = 2-a

Plugging this in the earlier equation gives 

\lim\limits_{x\to 1}\dfrac{\sqrt{ax+2-a}-\sqrt{x+1}}{x^2-1} = 4\sqrt{2}

rationalize numerator on left hand side and get
\lim\limits_{x\to 1}\dfrac{ax+2-a-x-1}{(x^2-1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\lim\limits_{x\to 1}\dfrac{a(x-1)-1(x-1)}{(x^2-1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\lim\limits_{x\to 1}\dfrac{(x-1)(a-1)}{(x-1)(x+1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\lim\limits_{x\to 1}\dfrac{a-1}{(x+1)(\sqrt{ax+2-a}+\sqrt{x+1})} = 4\sqrt{2}\\\dfrac{a-1}{(1+1)(\sqrt{a(1)+2-a}+\sqrt{1+1})} = 4\sqrt{2}\\\dfrac{a-1}{4\sqrt{2}} = 4\sqrt{2}
You can solve a
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