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)?
Best Answer
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: $$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}=\lim_{x\to 0}\frac{-\frac{7}{20}x^5+o(x^5)}{x^5}=-\frac{7}{20}$$