<s>[INST] Taylor limits with sine I'm having troubles calculating these two limits (I prefer to write a sigle question including both of them, instead of two different ones). 
This one 
$$\\ \lim_{x\rightarrow 0}  \frac{ x-\sin^2(\sqrt x)-\sin^2(x)} {x^2}  $$             I tried expanding with Taylors at different orders but the square root in the sine argument gave problems with the o() grades, leading for example to things like  this $$ o(\sqrt x^3)  $$ that does not seem easy to manage.  I tried rewriting the limit using $$ \sin^2(x) + \cos^2(x) = 1 $$, obtaining 
$$\\ \lim_{x\rightarrow 0}  \frac{ x-1+\cos^2(\sqrt x)-\sin^2(x)} {x^2}  $$
since the McLaurin expasion for the cosine looked a little better to me, but I keep obtaining a wrong result.
 The second limit in this one instead 
$$\\ \lim_{x\rightarrow 0}  \frac{ [\frac{1}{(1-x)} +e^x]^2 -4e^{2x} -2x^2} {x^3}  $$
and I happily wrote the McLaurin expansion stopping at the second order, that are 
$$(1-x)^{-1} = 1+x+\frac{x^2}{2} + o(x^2)$$
$$e^x = 1+x+\frac{x^2}{2} +o(x^2)$$
$$e^{2x} = 1 + 2x+2x^2+o(x^2)$$
doing the calculation leads to this 
$$\\ \lim_{x\rightarrow 0}\frac{\frac{-4}{3}x^3-2x^2}{x^3}$$ 
$$\\ \lim_{x\rightarrow 0}\frac{-4}{3} - \frac{2}{x}$$ and here I have problems understating what is going on. I don't know if x tends to 0 from right or left, so I can't evaluate the result (that would be, $$ +\infty$$ or $$ -\infty$$ ). Does this mean I did some wrong calculation? Do I need to take some more orders in the McLaurin expansion? 
 [/INST] If you're uncertain about $\sqrt{x}$, substitute $\sqrt{x}=t$, so the limit becomes
$$
\lim_{t\to0^+}\frac{t^2-\sin^2(t)-\sin^2(t^2)}{t^4}
$$
Of course we just need Taylor up to degree $4$:
\begin{align}
\sin^2(t)&=\left(t-\frac{t^3}{6}+o(t^3)\right)^2=t^2-\frac{t^4}{3}+o(t^4)\\[4px]
\sin^2(t^2)&=(t^2+o(t^2))^2=t^4+o(t^4)
\end{align}
Thus your limit is
$$
\lim_{t\to0^+}\frac{t^2-t^2+t^4/3-t^4+o(t^4)}{t^4}=-\frac{2}{3}
$$
For the second limit you should not stop at degree $2$ in the numerator, because the denominator has degree $3$. 

You have $$\left(\frac{1}{1-x}+e^x\right)^{\!2}=\left(2+2x+\frac{3}{2}x^2+\frac{7}{6}x^3+o(x^3)\right)^{\!2}=4+4x^2+8x+6x^2+\frac{14}{3}x^3+6x^3+o(x^3)$$ so the numerator becomes $$4+8x+10x^2+\frac{32}{3}x^3-4\left(1+2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{6}\right)-2x^2+o(x^3)=\frac{16}{3}x^3+o(x^3)$$

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