Computing $\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)$ I am trying to compute the following limit:
$$\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)$$
I have tried by rewriting it as $\lim_{x\to 0}\left(\frac{3x+(x^3-3)\tan(x)}{3x^4 \tan(x)}\right)$ and applying De l'Hopital's Rule but the expression quickly becomes unmanageable:
$$\lim_{x\to 0}\left(\frac{3x+(x^3-3)\tan(x)}{3x^4 \tan(x)}\right)\overset{H}{=}\lim_{x\to 0}\frac{3+(x^2 - 3) \sec^2(x) + 2 x \tan(x)}{3 x^3 (4 \tan(x) + x \sec^2(x))}$$
so I then tried by using the Maclaurin expression for $\tan(x)$:
\begin{align*}
\lim_{x\to 0}\left(\frac{\cot(x)}{x^3}-\frac{1}{x^4}+\frac{1}{3x^2}\right)&=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x-\tan(x)}{x^2\tan(x)}+\frac{1}{3}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{x-\left(x+\frac{x^3}{3}+\frac{2}{15}x^5\right)}{x\left(\frac{x^3}{3}+\frac{2}{15}x^5\right)}+\frac{1}{3}\right)\\
&=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}+\frac{1}{3}\right)=+\infty\cdot 0
\end{align*}
and I got an indeterminate form.
I am currently out of ideas so I would appreciate some help in figuring this out, thanks.

EDIT: It just occurred to me that
\begin{align*}
\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}+\frac{1}{3}\right)=\lim_{x\to 0}\frac{1}{x^2}\left(\frac{-\frac{1}{3}-\frac{2}{15}x^2+\frac{1}{3}+\frac{1}{9}x^2+\frac{2}{45}x^4}{1+\frac{x^2}{3}+\frac{2}{15}x^4}\right)\\ \lim_{x\to 0} \frac{1}{x^2}\left(\frac{-\frac{1}{45}x^2+\frac{2}{45}x^4}{1+\frac{x^2}{3}+\frac{2}{15}x^4}\right)=\lim_{x\to 0}\frac{-\frac{1}{45}+\frac{2}{45}x^2}{1+\frac{x^2}{3}+\frac{2}{15}x^4}=-\frac{1}{45}.
\end{align*}

The expression under limit can be written as $$\frac{3x-3\tan x+x^2\tan x}{3x^4\tan x}$$ Since $(\tan x) /x\to 1$ the desired limit is equal to the limit of $$\frac{3x-3\tan x+x^2\tan x}{3x^5}$$ Adding and subtracting $x^3$ in numerator we see that the expression can be rewritten as $$\frac{3x-3\tan x+x^3}{3x^5}+\frac{\tan x-x} {3x^3}$$ The last fraction tends to $1/9$ (via L'Hospital's Rule or Taylor expansion) and thus we need to evaluate the limit $L$ of first fraction above and get desired answer as $L+(1/9)$.
To get $L$ we can apply L'Hospital's Rule to get the expression $$\frac{1-\sec^2x+x^2} {5x^4} =\frac{x^2-\tan^2x}{5x^4}$$ which can be written as $$\frac{1}{5}\left(1+\frac{\tan x} {x} \right) \cdot\frac{x-\tan x} {x^3}$$ so that $$L=\frac{1}{5}\cdot 2\cdot \left(-\frac{1}{3}\right)=-\frac{2}{15}$$ The desired limit is $L+(1/9)=-1/45$. In the entire process L'Hospital's Rule has been used twice.
