# How do you evaluate limit of $\frac{\sqrt{a+x} - \sqrt a}{x\sqrt a(\sqrt{a+x})}$ when $x$ tends to $0$?

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.

• Please put dollar signs () around your math formulas to TeXify them, @zaidKnight. – Jose Arnaldo Bebita-Dris Jul 11 '18 at 15:28 • oh missed that .TY – zaidKnight Jul 11 '18 at 15:29 • is it right so? – Dr. Sonnhard Graubner Jul 11 '18 at 15:30 • yes. that looks proper now!! – zaidKnight Jul 11 '18 at 15:31 ## 3 Answers \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} • Oh! got it now. Thank you Siong. upvoted your answer but it wont show as my reputation is less than required. – zaidKnight Jul 11 '18 at 15:36 • no worries about reputation, glad you got it. – Siong Thye Goh Jul 11 '18 at 15:38 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 $$\frac{1}{\sqrt{a}}\lim_{x\to 0}\frac{f(x)-f(0)}{x}\times \lim_{x\to 0}\frac{1}{f(x)}=\frac{1}{\sqrt{a}}f'(0)\times \frac{1}{f(0)}=\frac{1}{2a^{3/2}}.$$ Easier solution: Writey=\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}