How to calculate this limit $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $? $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $
I tried L'Hopital, but it gets complicated.
I tried also this:
$$ \lim\limits_{x\rightarrow0}\frac{1-(1+x^2)\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
$$ \lim\limits_{x\rightarrow0}\frac{\sin^2{x}-x^2\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
, but I don't know what to do from here.
 A: I shall use this idea of yours:$$\lim_{x\to0}\frac{1-\sqrt{1+x^2}\cos(x)}{x^4}=\lim_{x\to0}\frac{1-(1+x^2)\cos^2(x)}{x^4\left(1+\sqrt{1+x^2}\cos(x)\right)}.$$Note that$$\lim_{x\to0}\left(1+\sqrt{1+x^2}\cos(x)\right)=2.$$So, what remains is to compute\begin{align}\lim_{x\to0}\frac{1-(1+x^2)\cos^2(x)}{x^4}&=\lim_{x\to0}\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}\\&=\lim_{x\to0}\frac{\sin(x)-x\cos(x)}{x^3}\times\lim_{x\to0}\frac{\sin(x)+x\cos(x)}x\\&=\frac13\times2\\&=\frac23,\end{align}and therefore your limit is equal to $\frac13$.
A: You can use Taylor expansion:
$$
\lim\limits_{x\to 0}\frac{1-\sqrt{1+x^2}\cos x}{x^4}=\lim\limits_{x\to 0}\frac{1-(1+x^2/2+\dots)(1-x^2/2+\dots)}{x^4}=\lim\limits_{x\to 0}\frac{x^4/3+\dots}{x^4}=1/3
$$
A: Another way.-Let $f(x)=\dfrac{1-\sqrt{1+x^2}\cos(x)}{x^4}$. Since $f(x)=f(-x)$ and is well defined for $|x|\gt0$ by prolongement by continuity we can put $f(0)=a$ where $a$ is precisely the asked limit. One has therefore
$$1-\sqrt{1+x^2}\cos(x)=ax^4$$ Now we have
$$1-(1+\frac{x^2}{2}-\frac{x^4}{8}+\cdots)(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots)=ax^4$$  The first coefficient in LHS is for $x^4$ and its value is equal to $\dfrac14+\dfrac18-\dfrac{1}{24}=\dfrac13$ so we have
$$\frac{x^4}{3}-\frac{x^6}{12}+\cdots=ax^4$$  from which the searched limit is $=\dfrac 13$
