# Solve limit containing (1/0) using L'Hopital's Rule

I was assigned a problem in my calculus 2 course that I can't seem to solve. We're going over indeterminate forms solved by L'Hopital's Rule.

$$\lim_{x\to \pi^-} \left( \frac{\csc(x)+x}{\tan(\frac{x}{2})} \right)$$

This is in the indeterminate form $\left[\frac{\infty}{\infty}\right]$, which is given, except I'm not sure why, as $\csc(\pi)$ is undefined. Is this because trig functions occur infinitely (within their range), e.g. $\sin(x) = 1$, occurs for infinite $x$'s?

After applying L'Hopital's Rule, I get:

$$\lim_{x\to \pi^-} \left( \frac{-\csc(x)\cot(x) + 1}{\frac{1}{2}\sec^2(\frac{x}{2})} \right)$$

However, I don't understand how I'm supposed to solve from here. No matter how many times we derive, $\csc(x)$ won't go away and I'm not aware of any trigonometric identities that can simplify the undefined division away. Am I missing something obvious?

• When you take $\lim_{x \to \pi^-} \csc(x)$, $\sin(x)$ is going to zero from positive values, so it goes to $+\infty$. The same is true of $\lim_{x \to \pi^-} \tan(x/2)$. – Ian Jan 28 '16 at 0:32

Essentially, you are considering $$\lim_{x\to\pi^-}\frac{(\sin x)^{-1}+x}{\frac{\sin(x/2)}{\cos(x/2)}} =\lim_{x\to\pi^-}\frac{1+x·\sin x}{2\sin^2(\frac x2)}$$ using $\sin(x)=2\sin(x/2)\cos(x/2)$, which (unfortunately) does not require l'Hopital to solve.