L'Hôpital's Rule for$\lim\limits_{x \to 0^+}\left( \frac 1 {x^2}-\frac 1 {\tan x } \right )$

Question

I'm struggling with the following problem: $$\lim\limits_{x \to 0^+}\left( \frac 1 {x^2}-\frac 1 {\tan x } \right )$$

Below is my work. But I'm stuck because my denominator equals 0. What did I do wrong? Also, a step through of this problem would be great.

oh really? so the answer of -1 / 0 is acceptable? the solution just goes to infinity ? — PineNuts0, Feb 11 '20 at 15:47
$\infty$ is a perfectly valid answer. You may verify this by plotting the graph in Desmos. — Sam, Feb 11 '20 at 16:07
This is also evident from the Laurent series of $\cot(z)$, indeed, around $0$, $\cot(z)$ behaves like $\frac1z$ — Maximilian Janisch, Feb 11 '20 at 16:20

score 2 · Answer 1 · answered Feb 12 '20 at 03:57

2

I'd suggest a different approach using

$\frac{\tan x}{x}\stackrel{x \to 0}{\longrightarrow}1$

$$\left( \frac 1 {x^2}-\frac 1 {\tan x } \right ) = \frac 1{x^2}\underbrace{\left( 1-\underbrace{x\frac x {\tan x }}_{\stackrel{x \to 0}{\longrightarrow}0\cdot 1=0} \right )}_{\stackrel{x \to 0}{\longrightarrow}1}\stackrel{x \to 0}{\longrightarrow}+\infty$$

answered Feb 12 '20 at 03:57

trancelocation

32,243

Simplest approach. +1 – Paramanand Singh Feb 12 '20 at 06:12

score 1 · Answer 2 · answered Feb 11 '20 at 16:24

As the commenters mention, $\lim_{x\to 0+} \big( \frac1{x^2}-\frac1{\tan x} \big)=+\infty$.

l'Hôpital's rule says that (under suitable conditions) $\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}$. In other words, it transforms one problem into another problem; if we can solve the new problem, then the old problem has the same solution.

If we encountered the problem $\lim_{x\to 0+} \frac{\sec^2 x-2x}{x^2\sec^2 x+2x\tan x}$ in the wild, we would see that the numerator tends to $1$ while the denominator tends to $0$ through positive values, and hence the limit is $+\infty$. And we wouldn't worry that the denominator tended to $0$ because we know limits act that way sometimes.

So the same is true if we encounter this problem after an application of l'Hôpital's rule. It's a feature, not a bug!

score 1 · Answer 3 · answered Feb 11 '20 at 16:37

It would be very easy to avoid the L'Hospital Rule. Anyway if you want to use it, you can observe that $$ f(x) = \frac{1} {{x^2 }} - \frac{1} {{\tan x}} = \frac{{\tan x - x^2 }} {{x^2 \tan x}} \sim \frac{{\tan x - x^2 }} {{x^3 }} = \frac{{F(x)}} {{G(x)}},\,\,\,\,\left( {x \to 0} \right) $$ Now $$ \frac{{F'(x)}} {{G'(x)}} = \frac{{1 + \tan ^2 x - 2x}} {{3x^2 }} $$ and since $$ \mathop {\lim }\limits_{x \to 0} \frac{{1 + \tan ^2 x - 2x}} {{3x^2 }} = + \infty $$ ad all the hypothesis for de L'Hospital Rule are satisfied, you have that $$ \mathop {\lim }\limits_{x \to 0} \frac{{\tan x - x^2 }} {{x^3 }} = + \infty $$ so $$ \mathop {\lim }\limits_{x \to 0} f(x) = + \infty $$ in particular your limit is $+\infty$

L'Hôpital's Rule for$\lim\limits_{x \to 0^+}\left( \frac 1 {x^2}-\frac 1 {\tan x } \right )$

3 Answers3