Evaluate $\lim_{x\to 0} {x\sqrt{y^2-(y-x)^2}\over (\sqrt{8xy-4x^2}-\sqrt{8xy})^3}$ Could any one tell me how to solve this limit?I tried doing rationalizations and all others, but failed. Thank you
$$\lim_{x\to 0} {x\sqrt{y^2-(y-x)^2}\over (\sqrt{8xy-4x^2}-\sqrt{8xy})^3}$$
 A: If $y\ne0$ then both numerator and denominator have a factor $x\sqrt{x}$.  After cancelling that, the limit of the numerator is $\sqrt{2y}$ and the limit of the denominator is $0$, so your question has no limit.
If $y=0$ then the expression is $ix^2/(2ix)^3$ which has no limit either.
A: Taylor expansion seems to be the easiest way.
Consider $$A=\sqrt{y^2-(y-x)^2}=\sqrt{2xy-x^2}=\sqrt{2x}\sqrt{1-\frac{x}{2y}}$$ Now, use Taylor series of $\sqrt {1+z}$ and replace $z$ by $-\frac{x}{2y}$; you will get $$A=\sqrt{2} \sqrt{x} \sqrt{y}-\frac{x^{3/2}}{2 \left(\sqrt{2}
   \sqrt{y}\right)}+O\left(x^{5/2}\right)$$ Do something similar with $$B=\sqrt{8xy-4x^2}-\sqrt{8xy}$$ $$B=-\frac{x^{3/2}}{\sqrt{2} \sqrt{y}}+O\left(x^{5/2}\right)$$ Now you have to consider $\frac{x A}{B^3}$.
I am sure that you can take from here and arrive to Michael's conclusion.
In fact, you could show that $$\lim_{x\to 0} {x\sqrt{y^2-(y-x)^2}\over (\sqrt{8xy-4x^2}-\sqrt{8xy})^n}$$ only exist if $n\leq 1$.
A: Er, similar to the Michael's idea, I try to utilize the rationalizations to calculate this limit
  $$\lim_{x\to 0} {x\sqrt{y^2-(y-x)^2}\over (\sqrt{8xy-4x^2}-\sqrt{8xy})^3}$$
$$=\frac{1}{2} \lim_{x\to 0} {x\sqrt{2xy-x^{2}}\over (\sqrt{2xy-x^2}-\sqrt{2xy})^3}$$
$$=\frac{1}{2} \lim_{x\to 0} {\sqrt{2xy-x^{2}}(\sqrt{2xy-x^2}+\sqrt{2xy})^3\over -x^{5}}$$
It is not hard to see the highest power of $x$ in numerator is only 4. So this limit  tends to infinity.
