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I tried rationalization method where we multiply both the numerator and denominator with appropriate opposite factor of numerator. But I could only get $\frac{1}{\sqrt {a^2+ax}(2 \sqrt {a + \sqrt x)}}$. But the final solution in textbook says it should be $\frac{1}{2a^{3/2}}$. Please help.

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    $\begingroup$ Please put dollar signs ($) around your math formulas to TeXify them, @zaidKnight. $\endgroup$ – Jose Arnaldo Bebita-Dris Jul 11 '18 at 15:28
  • $\begingroup$ oh missed that .TY $\endgroup$ – zaidKnight Jul 11 '18 at 15:29
  • $\begingroup$ is it right so? $\endgroup$ – Dr. Sonnhard Graubner Jul 11 '18 at 15:30
  • $\begingroup$ yes. that looks proper now!! $\endgroup$ – zaidKnight Jul 11 '18 at 15:31
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\begin{align}\lim_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}\sqrt{a+x}} &= \lim_{x \to 0}\frac{a+x-a}{x\sqrt{a}\sqrt{a+x}(\sqrt{a+x}+\sqrt{a})}\\&=\lim_{x\to 0}\frac{1}{\sqrt{a}\sqrt{a+x}(\sqrt{a+x}+\sqrt{a})}\\ &= \frac{1}{\sqrt{a}\sqrt{a}(2\sqrt{a})}\\ &=\frac{1}{2a^\frac32}\end{align}

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  • $\begingroup$ Oh! got it now. Thank you Siong. upvoted your answer but it wont show as my reputation is less than required. $\endgroup$ – zaidKnight Jul 11 '18 at 15:36
  • $\begingroup$ no worries about reputation, glad you got it. $\endgroup$ – Siong Thye Goh Jul 11 '18 at 15:38
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Another solution. Put $$ f(x)=\sqrt{x+a}\implies f'(x)=\frac{1}{2\sqrt{x+a}}. $$ by the chain rule. Then the limit can be expressed as $$ \frac{1}{\sqrt{a}}\lim_{x\to 0}\frac{f(x)-f(0)}{x}\times \lim_{x\to 0}\frac{1}{f(x)}=\frac{1}{\sqrt{a}}f'(0)\times \frac{1}{f(0)}=\frac{1}{2a^{3/2}}. $$

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Easier solution: Write $y=\sqrt{x+a}$ and $b=\sqrt{a}$, then

\begin{align}\lim_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}\sqrt{a+x}} &= \lim_{y \to b}\frac{y-b}{by(y^2-b^2)}\\&=\lim_{y \to b}\frac{1}{by(y+b)}\\ &= \frac{1}{2b^3}=\frac{1}{2a\sqrt{a}}\end{align}

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