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

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.

• It's ok. It means your limit equals $\infty$. – LHF Feb 11 at 15:42
• oh really? so the answer of -1 / 0 is acceptable? the solution just goes to infinity ? – PineNuts0 Feb 11 at 15:47
• $\infty$ is a perfectly valid answer. You may verify this by plotting the graph in Desmos. – Sam Feb 11 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 at 16:20

I'd suggest a different approach using

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

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

• Simplest approach. +1 – Paramanand Singh Feb 12 at 6:12

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!

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 $$+\infty$$