<s>[INST] 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.
 [/INST] 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 $=\dfrac 13$
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