Finding limit using L'Hospital's rule.

Question

Find $$\lim_{x\to 0}{\left(\cot^2x- {1\over x^2}\right)}$$

This is $${\infty - \infty} \ \ \ \text{form}$$ so I converted it into $$\lim_{x\to 0} \left({x^2 - \tan^2x} \over {x^2 \tan^2x}\right)$$ i.e. in $${0\over 0} \ \ \text{form}$$
But after applying L 'Hospital's rule the problem gets complicated.

Answer 1

Hint:

We have $$\dfrac{x-\tan x}{x^3}\dfrac{x+\tan x}x\left(\dfrac x{\tan x}\right)^2$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion

and $\dfrac{x+\tan x}x=1+\dfrac{\tan x}x$

Answer 2

The limit, as given is in the form $\infty-\infty$. Your idea of transforming it using the tangent is good, but the next form $$ \lim_{x\to0}\frac{x^2-\tan^2x}{x^2\tan^2x}\tag{1} $$ might require four applications of l'Hôpital. And the fourth derivative of $x^2\tan^2x$ is a terrible mess.

However, if you recall that $\lim_{x\to0}\frac{\tan x}{x}=1$, you can be certain that, if $$ \lim_{x\to0}\frac{x^2-\tan^2x}{x^4}\tag{2} $$ exists, then it is equal to the limit in $(1)$. This simplifies at least the derivatives in the denominator. Let's go on with the first application \begin{align} (2) &= \lim_{x\to0}\frac{2x-2\tan x(1+\tan^2 x)}{4x^3}\tag{3} \\[4px] &=\frac{1}{2}\lim_{x\to0}\frac{x-\tan x-\tan^3x}{x^3} \\[4px] &=\frac{1}{2}\lim_{x\to0}\frac{1-(1+\tan^2x)-3\tan^2x(1+\tan^2x)}{3x^2} \\[4px] &=-\frac{1}{6}\lim_{x\to0}\frac{\tan^2x(4+3\tan^2x)}{x^2} \end{align} and now we can actually stop, because of the already used known limit. You can end up.

Answer 3

Hint.

Considering $\tan x = x+\frac{x^3}{3}+O(x^4)$ we have

$$ \lim_{x\to 0}\left(\frac{x+\tan x}{x\tan x}\right)\left(\frac{x-\tan x}{x\tan x}\right) = -\lim_{x\to 0}\frac{x \left(\frac{x^3}{3}+2 x\right)}{3 \left(\frac{x^3}{3}+x\right)^2} $$

Answer 4

As an alternative by Taylor's expansion

• $\tan x=x+\frac13x^3+o(x^3)$

we have

$$\frac{x^2 - \tan^2x} {x^2 \tan^2x}=\frac{x^2 - \left(x+\frac13x^3+o(x^3)\right)^2} {x^2 \left(x+\frac13x^3+o(x^3)\right)^2}=\frac{x^2-x^2-\frac23x^4+o(x^4)}{x^4+o(x^4)}=\frac{-\frac23+o(1)}{1+o(1)}\to -\frac23$$

Answer 5

You can rewrite things as

$${x^2-\tan^2x\over x^2\tan^2x}=\left(x-\tan x\over x^3 \right)\left(x+\tan x\over x \right)\left(x\over\tan x \right)^2$$

and apply L'Hopital to each piece separately. In particular,

$$\lim_{x\to0}\left(x\over\tan x \right)^2=\left(\lim_{x\to0}{x\over\tan x}\right)^2=\left(\lim_{x\to0}{1\over\sec^2x}\right)^2=1$$

after which we have

$$\lim_{x\to0}\left(x-\tan x\over x^3 \right)=\lim_{x\to0}\left(1-\sec^2x\over 3x^2 \right)=\lim_{x\to0}\left(-\tan^2x\over3x^2\right)=-{1\over3}\left(\lim_{x\to0}{\tan x\over x}\right)^2=-{1\over3}$$

(making use of a the trig identity $\tan^2x=\sec^2x-1$). Putting this together with

$$\lim_{x\to0}\left(x+\tan x\over x \right)=\lim_{x\to0}\left(1+\sec^2x\over 1 \right)=2$$

we get

$$\lim_{x\to0}\left(x^2-\tan^2x\over x^2\tan^2x\right)=-{1\over3}\cdot2\cdot1=-{2\over3}$$

The take-home lesson here is that some seemingly messy limit calculations can be carried out by breaking things into nice clean pieces. The trick is learning to recognize which pieces are nice and clean.

Remark: It's not really necessary (and some would even say it's inappropriate) to use L'Hopital to get ${\tan x\over x}\to1$ or ${x+\tan x\over x}\to2$ as $x\to0$. These limits are really just the definition of the derivative $f'(0)$ for $f(x)=\tan x$ and $f(x)=x+\tan x$, respectively.