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

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 $A-(1/6)=-7/20$.
