# Finding the limit using McLaurin series and/or "little o" method.

Another mathematical analysis question to do with limits. Tried a few approaches, but I'm not getting the correct answer. Question:Find the limit of:

$\lim_{x\rightarrow0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^{2}x}$

Now, we know as x approaches zero, the following “little o” approximations apply:

$\cos x=1-\frac{x^{2}}{2}+o(x)$ and $\sin x=x+o(x)$

So far I've tried substituting these approximations into the limit and using the McLaurin series expansion (the derivation of which I'm not including for sake of brevity, but I can assure you they were correct), however that just gives me a value of “1” for $\cos x$, which gives me a limit of zero. The actual answer should be $\frac{1}{12}$.

Your starting point $$\cos x \simeq 1-\frac{x^{2}}{2}$$ $$\sin x \simeq x$$ is good. Now, you have to develop the power of the numerator, remembering that, when $y$ is small $$(1+y)^n \simeq 1+ny$$ So $$\sqrt {\cos x}=(\cos x)^{1/2} \simeq 1-\frac{x^{2}}{4}$$ $$\sqrt[3]{\cos x}=(\cos x)^{1/3} \simeq 1-\frac{x^{2}}{6}$$
• As I said, when $y$ is small compared to $1$, $(1+y)^n \simeq 1+ny$. Just apply to $(\cos x)^{1/2}$ and $(\cos x)^{1/3}$. May 3 '14 at 7:59