Evaluate $\lim_{x\to0}\frac{\sqrt{a^2+x}-|a|}{\sin^a2x\ln\cos x}$

Question

I have just learned limits and series by myself, but I'm stuck with this limit: $$\lim_{x\to0}\frac{\sqrt{a^2+x}-|a|}{\sin^a2x\ln\cos x}$$ I would like to evaluate that limit with $a\in\mathbb R$. I would also like to understand in detail the steps involved in order to solve that. Thanks.

Answer 1

HINT

We have that

$$\ln(\cos x)=\ln(1-x^2/2+o(x^2))=-\frac12x^2+o(x^2)$$

$$\sqrt { { a }^{ 2 }+x }=|a|\sqrt { 1+x/a^2 }=|a|\left(1+\frac{x}{2a^2}+o(x)\right)=|a|+\frac{x}{2|a|}+o(x)$$

therefore

$$ \frac { \sqrt { { a }^{ 2 }+x } -|a| }{ { \sin }^{ a }(2x)\ln(\cos x) }= \frac { (2x)^a}{ { \sin }^{ a }(2x) } \frac { \frac{x}{2|a|}+o(x) }{ (2x)^a\left(-\frac12x^2+o(x^2)\right) }$$

from here we can conclude that for $a=-1$ the limit exists finite.

What about $a>-1$ and $a<-1$?

Answer 2

For real $a,$

$$\lim_{x\to0}\frac{\sqrt{a^2+x}-|a|}{\sin^a2x\ln\cos x}$$

$$=-\dfrac2{2^a}\lim_{x\to0}\dfrac{a^2+x-a^2}x\cdot\dfrac1{\lim_{x\to0}(\sqrt{a^2+x}+|a|)}\cdot\lim_{x\to0}\dfrac{-\sin^2x}{\ln(1-\sin^2x)}\left(\lim_{x\to0}\frac{2x}{\sin2x}\right)^a\left(\lim_{x\to0}\dfrac{\sin x}x\right)^2\dfrac1{\lim_{x\to0}x^{a-2}}$$

$$=-\dfrac2{2^a2|a|}\cdot\dfrac1{\lim_{x\to0}x^{a-2}}$$