# How do you use L'hospital's rule to find the limit lim_(x->oo)xsin(pi/x) ?

I assume that the limit is at $+ \infty$ (otherwise the same arguments can be easily adapted to the $- \infty$ case). So ${\lim}_{x \to + \infty} x \sin \left(\frac{\pi}{x}\right)$.

In this case you have an indeterminate form of type $+ \infty \cdot 0$. In fact, the first factor is simply $x \to + \infty$ and the second one is $\sin \left(\frac{\pi}{x}\right) \to 0$, because $\frac{\pi}{x} \to 0$ when $x \to + \infty$ and $\sin \left(0\right) = 0$.
There are other (easier?) ways to solve this (see below) than the "brutal" application of L'Hôpital's rule, which is sometimes troublesome. Anyway, your question is precise so let's go ahead.

L'Hôpital's rule states the following: if ${x}_{0} \in m a t h \boldsymbol{R} \cup \left\{\pm \infty\right\}$ and $f$,$g$ are two functions such that ${\lim}_{x \to {x}_{0}} f \frac{x}{g} \left(x\right)$ is an indeterminate form $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$, then only if the limit ${\lim}_{x \to {x}_{0}} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ exists, the following equality holds:
${\lim}_{x \to {x}_{0}} f \frac{x}{g} \left(x\right) = {\lim}_{x \to {x}_{0}} \frac{f ' \left(x\right)}{g ' \left(x\right)}$.
Notice the specification in bold. That's the "troublesomeness" of the rule: if the limit of the ratio $\frac{f ' \left(x\right)}{g ' \left(x\right)}$ doesn't exist, we can't conclude that the limit of $f \frac{x}{g} \left(x\right)$ doesn't exist too. Here you can find an example of limit for which the rule isn't valid.

Back to our limit: we have to manipulate the expression so that the indeterminate form $+ \infty \cdot 0$ becomes $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$. How can a product be turned into a fraction?
The following equality holds:
$x = \frac{1}{\frac{1}{x}}$
So we can use this in the former limit and get:
${\lim}_{x \to + \infty} x \sin \left(\frac{\pi}{x}\right) = {\lim}_{x \to + \infty} \frac{\sin \left(\frac{\pi}{x}\right)}{\frac{1}{x}}$
Now $\sin \left(\frac{\pi}{x}\right) \to 0$ and $\frac{1}{x} \to 0$ as $x \to + \infty$, so we got the case $\frac{0}{0}$. Now we can compute derivatives and hope that the limit of their ratio exists:
$f \left(x\right) = \sin \left(\frac{\pi}{x}\right) \implies f ' \left(x\right) = - \frac{\pi}{x} ^ 2 \cos \left(\frac{\pi}{x}\right)$
$g \left(x\right) = \frac{1}{x} \implies g ' \left(x\right) = - \frac{1}{x} ^ 2$
${\lim}_{x \to + \infty} \frac{- \frac{\pi}{x} ^ 2 \cos \left(\frac{\pi}{x}\right)}{- \frac{1}{x} ^ 2} = {\lim}_{x \to + \infty} \pi \cos \left(\frac{\pi}{x}\right) = \pi \cdot \cos \left(0\right) = \pi$
The limit exists, so by L'Hôpital's rule we conclude that
${\lim}_{x \to + \infty} x \sin \left(\frac{\pi}{x}\right) = \pi$

Notice that this limit is easier to solve if you multiply the function by $1 = \frac{\frac{\pi}{x}}{\frac{\pi}{x}}$:
${\lim}_{x \to + \infty} x \cdot \frac{\frac{\pi}{x}}{\frac{\pi}{x}} \cdot \sin \left(\frac{\pi}{x}\right) = {\lim}_{x \to + \infty} \pi \cdot \sin \frac{\frac{\pi}{x}}{\frac{\pi}{x}}$. Now we can apply the following notable limit:
${\lim}_{\epsilon \to 0} \sin \frac{\epsilon}{\epsilon} = 1$
Here we have that $\epsilon = \frac{\pi}{x} \to 0$ if $x \to + \infty$. So we get that ${\lim}_{x \to + \infty} \pi \cdot \sin \frac{\frac{\pi}{x}}{\frac{\pi}{x}} = \pi \cdot 1 = \pi$