# How to calculate $\lim_{x\rightarrow 0^+}\frac{x}{tg^2x}$ without De L'Hopital?

I would like to calculate the following limit: $$\lim_{x\rightarrow 0^+}\frac{x}{tg^2x}$$ without using De L'Hopital rule. I know that as $$x\rightarrow 0^+$$, it gives an indeterminate form of the type $$\frac{0}{0}$$. Therefore using De L'Hopital seems a reasonable choice, but I was wondering if we could calculate the limit using some kind of algebraic manipulation.

• I do not know of any simple algebraic manipulation but without using De L'Hôpital, remember that $\tan x \underset{x\to 0^+}{\sim}x$ is just enough to conclude here. – Bill O'Haran May 28 '19 at 13:59

$$y = \lim_{x\to0^+}\frac{x}{\tan^2x} = \lim_{x\to0^+}\frac{x^2}{x\tan^2x} = \lim_{x\to0^+}\frac{1}{x} = \infty$$
As, $$\lim_{x\to0^-}\frac{1}{x} = -\infty$$ and $$\lim_{x\to0^+}\frac{1}{x} = +\infty$$, the limit doesn't exist at 0
If you want more than the limit itself, use Taylor expansions $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ $$\frac x {\tan^2(x)}=\frac x {\left(x+\frac{x^3}{3}+O\left(x^5\right)\right)^2}=\frac x {x^2+\frac{2 x^4}{3}+O\left(x^6\right)}=\frac{1}{x}-\frac{2 x}{3}+O\left(x^3\right)$$ which shows the limit and also how it is approached.