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? 
 A: 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 $$4+8x+10x^2+\frac{32}{3}x^3-4\left(1+2x+\frac{(2x)^2}{2}+\frac{(2x)^3}{6}\right)-2x^2+o(x^3)=\frac{16}{3}x^3+o(x^3)$$

A: First limit
Note that
$$sin^2 \sqrt x=\left(\sqrt x-\frac{x\sqrt x}{6}+o(x\sqrt x)\right)^2=x-\frac{x^2}{3}+o(x^2)$$
$$\sin^2 x=x^2+o(x^2)$$
thus
$$\frac{ x-\sin^2 \sqrt x -\sin^2 x} {x^2}=\frac{ x-x+\frac{x^2}{3}-x^2+o(x^2)} {x^2}=\frac{ \frac{-2x^2}{3}+o(x^2)} {x^2}=-\frac23+o(1)\to-\frac23$$
A: Second limit
Note that:
$$(1-x)^{-1}=1+x+x^2+x^3+o(x^3)$$
$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)$$
$$({(1-x)^{-1}} +e^x)^2=\left(1+x+x^2+x^3+1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)\right)^2=\left(2+2x+\frac{3x^2}{2}+\frac{7x^3}{6}+o(x^3)\right)^2=4+4x^2+8x+6x^2+6x^3+\frac{14x^3}{3}+o(x^3)=4+8x+10x^2+\frac{32x^3}{3}+o(x^3)$$
$$e^{2x}=1+2x+\frac{4x^2}{2}+\frac{8x^3}{6}+o(x^3)=1+2x+2x^2+\frac{4x^3}{3}+o(x^3)$$
thus
$$\frac{ ({(1-x)^{-1}} +e^x)^2 -4e^{2x} -2x^2} {x^3}=\frac{4+8x+10x^2+\frac{32x^3}{3}-4-8x-8x^2-\frac{16x^3}{3}-2x^2+o(x^3)} {x^3}=\frac{\frac{16x^3}{3}+o(x^3)} {x^3}=\frac{16}{3}+o(1)\to\frac{16}{3}$$
