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(pi / x)#.

In this case you have an indeterminate form of type #+infty * 0#. In fact, the first factor is simply #x to +infty# and the second one is #sin(pi/x) to 0#, because #pi/x to 0# when #x to +infty# and #sin(0)=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 mathbb{R} cup {pm infty}# and #f#,#g# are two functions such that #lim_{x to x_0} f(x)/g(x)# is an indeterminate form #0/0# or #pm infty/infty#, then **only if the limit #lim_{x to x_0} {f'(x)} / {g'(x)}# exists**, the following equality holds:

#lim_{x to x_0} f(x) / g(x) =lim_{x to x_0} {f'(x)} / {g'(x)} #.

Notice the specification in bold. That's the "troublesomeness" of the rule: if the limit of the ratio #[f'(x)}/{g'(x)}# doesn't exist, we can't conclude that the limit of #f(x)/g(x)# 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 * 0# becomes #0/0# or #pm infty / infty#. How can a product be turned into a fraction?

The following equality holds:

#x=1/{1/x}#

So we can use this in the former limit and get:

#lim_{x to +infty} x sin (pi/x)=lim_{x to +infty} {sin (pi/x)}/{1/x}#

Now #sin(pi/x) to 0# and #1/x to 0# as #x to +infty#, so we got the case #0/0#. Now we can compute derivatives and hope that the limit of their ratio exists:

#f(x)=sin(pi/x) => f'(x)=-pi/x^2 cos(pi/x)#

#g(x)=1/x => g'(x)=-1/x^2#

#lim_{x to +infty} {-pi/x^2 cos(pi/x)}/{-1/x^2}=lim_{x to +infty} pi cos(pi/x)=pi *cos(0)=pi#

The limit exists, so by L'Hôpital's rule we conclude that

#lim_{x to +infty} x sin (pi/x)=pi#

Notice that this limit is easier to solve if you multiply the function by #1=(pi/x)/(pi/x)#:

#lim_{x to +infty} x *(pi/x)/(pi/x) *sin (pi/x) =lim_{x to +infty} pi*sin(pi/x)/(pi/x)#. Now we can apply the following notable limit:

#lim_{epsilon to 0} sin(epsilon) / epsilon = 1#

Here we have that #epsilon=pi/x to 0# if #x to +infty#. So we get that #lim_{x to +infty} pi*sin(pi/x)/(pi/x)=pi*1=pi#