Evaluating $\lim_{x\to 0}\frac{\cos(\sin x)-(1+x^2)^{\frac{-1}{2}}}{x^4}$ $$L=\lim_{x\to 0}\frac{\cos(\sin x)-(1+x^2)^{\frac{-1}{2}}}{x^4}$$
To Evaluate the limit I used few terms of taylor series:
$\cos(\sin x)=1-\frac{\sin^2 x}2+\cdots$
$(1+x^2)^{\frac{-1}2}=1-\frac12x^2+\cdots$
$$L=\lim_{x\to 0}\frac{1-\frac{\sin^2 x}{2}-1+\frac{x^2}{2}}{x^4}=\lim_{x\to0}\frac{x^2-\sin^2 x}{2x^4}$$
$\sin^2 x=(x-\frac{x^3}{6}+O(x^4))^2=x^2-\frac{x^4}{3}+O(x^5)$
So we have:$$\lim_{x\to 0}\frac{x^2-x^2+\frac{x^4}{3}}{2x^4}=\frac{1}{6}$$
But the right answer is $\frac{-1}6$. Why my final answer is wrong?
This is a standard application of Taylor series, hence Community Wiki.
We have $\cos x=1-\frac12x^2+\frac1{24}x^4-\cdots$ and $\sin x=x-\frac16x^3+\cdots$, so
$$
\begin{aligned}
\cos (\sin x)&amp;=1-\frac12\sin^2 x+\frac1{24}\sin^4x+\cdots\\
&amp;=1-(\frac12x^2-\frac16x^4+O(x^6))+\frac1{24}(x^4+O(x^6))\\
&amp;=1-\frac12x^2+\frac5{24}x^4+O(x^6).
\end{aligned}
$$
Similarly (use the binomial series) the Taylor series of
$$
1/\sqrt{1+x^2}=1-\frac12x^2+\frac9{24}x^4+O(x^6).
$$
Therefore
$$
\cos(\sin x)-\frac1{\sqrt{1+x^2}}=-\frac16x^4+
O(x^6).
$$
From this it follows immediately that the limit is equal to $-1/6$.
Everywhere $O(x^6)$ stands for terms of degree six or higher.

The problem with the attempt in the original post is that in the first step some terms of degree four were neglected. In the end we cancel a factor of $x^4$, so this was a mistake.
Write  $\cos(\sin x)-(1+x^2)^{1/2}$ as $\cos(\sin x)-\cos x+\cos x-(1+x^2)^{1/2}$
Then,
$$\cos(\sin x)-\cos x=2\sin\dfrac{x-\sin x}2\sin\dfrac{x+\sin x}2$$
Now use L'Hospital's or  Are all limits solvable without L&#39;H&#244;pital Rule or Series Expansion to find $$\lim_{x\to0}\dfrac{x-\sin x}{x^3}=\dfrac16$$  so that $$\lim_{x\to0}\dfrac{\cos(\sin x)-\cos x}{x^4}=\dfrac24\lim_{x\to0}\dfrac{\sin\dfrac{x-\sin x}2}{\dfrac{x-\sin x}2}\lim_{x\to0}\dfrac{\sin\dfrac{x+\sin x}2}{\dfrac{x+\sin x}2}\lim_{x\to0}\dfrac{x-\sin x}{x^3}\lim_{x\to0}\dfrac{x+\sin x}x$$
Finally $$\cos x-(1+x^2)^{-1/2}$$
$$=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots-\left(1-\dfrac{x^2}2+x^4\cdot\dfrac{\left(-\dfrac12\right)\left(-\dfrac12-1\right)}{2!}+\cdots\right)$$
$$=x^4\cdot\left(\dfrac1{4!}-\dfrac38\right)+\text{terms containing higher exponents of }x$$