$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$ 
Evaluate $$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$$

Wolfram alpha gives $\dfrac{-7}{20}$.
Here is my work
For $x \to 0 $
$\tan x \sim x$
$\sin(x+\tan x) \sim \sin2x$
$\sin2x \sim 2x$ (I wasn't sure about this but I evaluated the limit  and got $1$.)
$\sin(x+\tan x) \sim 2x$
$\sin(x+\tan x)-2x\cos x \sim 2x(1-\cos x)$
$(1-\cos x) \sim x$
So I got that
$\dfrac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2} \sim \dfrac{3x}{x(\sin x^2)^2}$
which means I got $\infty$.
Did make a mistake somewhere?
And if so can this problem be solved by this approach or I need to try something else (like l'hopitals or something)?
 A: You have messed up with asymptotic relations. You must be really really careful when you use them.
In this limit, we know from the denominator that we have to search for a $5$ degree expansion. Namely:
$$x\cdot(\sin(x^2))^2\,\,\sim\,\, x^5$$
Now, we set things ready for Tyalor-MacLaurin series:
$$\sin(t)=t-\frac{1}{6}t^3+\frac{1}{120}t^5+o(t^5)$$
$$\tan(t)=t+\frac{1}{3}t^3+\frac{2}{15}t^5+o(t^5)$$
$$\cos(t)=1-\frac{1}{2}t^2+\frac{1}{24}t^4+o(t^4)$$
Now, we can work on t numerator:
$$\sin(x+\tan(x))-2x\cos(x)=\sin\left(2x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)\right)-2x\left(1-\frac{1}{2}x^2+\frac{1}{24}x^4+o(x^4)\right)=2x+\frac{1}{3}x^3+\frac{1}{120}x^5-\frac{1}{6}\left(2x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)\right)^3+\frac{1}{120}\left(2x+\frac{1}{3}x^3+\frac{2}{15}x^5+o(x^5)\right)^5-2x+x^3-\frac{1}{12}x^5+o(x^5)=-\frac{7}{20}x^5+o(x^5)$$
Note that, here, I haven't done all the calculations in order to exapand the $3$ and $5$ powers of quadrinomials, but I have kept only the most significant (grade $1$, $3$ and $5$).
So:
$$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}=\lim_{x\to 0}\frac{-\frac{7}{20}x^5+o(x^5)}{x^5}=-\frac{7}{20}$$
A: You have used the asymptotic relations like $\sin  x\sim x$ in an incorrect fashion. The meaning of such relation is $$\lim_{x\to 0} \frac{\sin x} {x} =1$$ It does not mean that you can replace $\sin x$ with $x$ in any limit evaluation as $x\to 0$.  I have explained this in detail in this answer of mine.
The simplest approach here is to use Taylor series expansions, but I usually avoid multiplication, division and composition of Taylor series. The approach below uses a little algebraic manipulation before applying Taylor series expansions.
Using L'Hospital's Rule or Taylor series one can establish the following limits
\begin{align}
\lim_{x\to 0}\frac{\sin x-x} {x^3}&=-\frac{1}{6}\tag{1}\\
\lim_{x\to 0}\frac {\tan x-x} {x^3}&=\frac{1}{3}\tag{2}
\end{align}
Replacing $x$ with $\tan x$ in $(1)$ we get $$\lim_{x\to 0} \frac{\sin\tan x - \tan x} {\tan^3x}=-\frac {1}{6}$$ Multiplying this with the cube of $$\lim_{x\to 0}\frac{\tan x} {x} =1$$ we get $$\lim_{x\to 0}\frac{\sin\tan x-\tan x} {x^3}=-\frac{1}{6}$$ Adding the above to equation $(2)$ we get $$\lim_{x\to 0} \frac{\sin\tan x-x}{x^3}=\frac{1}{6}\tag{3}$$ Let $$u=\sin\tan x-x, v=\tan x-x\tag{4}$$ and then we can rewrite the equations $(2),(3)$ as $$u/x^3\to 1/6,v/x^3\to 1/3\tag{5}$$ as $x\to 0$.
Using the identity $$\sin(\tan x+x)+\sin(\tan x-x) =2\sin\tan x\cos x$$ the numerator of the fraction under limit in question can be written as $$2u\cos x-\sin v$$ and the denominator can be replaced by $x^5$ (justify this replacement using the limit $\lim_{x\to 0}(\sin x) /x=1$). Thus we need to evaluate the limit of expression $$\frac{2u\cos x-\sin v} {x^5}$$ The above can be written as $$\frac {2u(\cos x-1)}{x^5}+\frac{2u-v}{x^5}+\frac{v-\sin v} {x^5}\tag {6}$$ The first fraction above can be written as $$\frac{2u}{x^3}\cdot \frac{\cos x-1}{x^2}$$ so that it tends to $$2(1/6)(-1/2)=-1/6$$ via $(5)$ and the last fraction in $(6)$ can be rewritten as $$\frac{v-\sin v} {v^3}\cdot\left(\frac {v} {x^3}\right)^3\cdot x^4$$ which tends to $$(1/6)(1/3)^3\cdot 0^4=0$$ The middle fraction in equation $(6)$ tends to a limit, say $A$, as shown below. And therefore the desired limit is $A-(1/6)$.
The middle fraction in equation $(6)$ is $$\frac{2\sin\tan x-x-\tan x} {x^5}$$ Using the substitution $t=\tan x$ and the limit $$\lim_{t\to 0}\frac{\arctan t} {t} =1$$ the above is reduced to $$\frac{2\sin t-t-\arctan t} {t^5}$$ and using Taylor expansions we see that it tends to $A=-11/60$. The desired limit is then $A-(1/6)=-7/20$.
