How to use L Hospital's rule for $\lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) $ I want to find the following limit using L Hospital's rule:
 $$ \lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) $$ 
I know that this can be solved using squeezed theorem from Cal 1:
$$ 0 < \sqrt{x}\sin( \frac{1}{x} ) < \frac{1}{x} $$ 
since $0 < \sin( \frac{1}{x}) < \frac{1}{x} $.
What I have done so far is trying to convert it to fraction form 
$$ \lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) = \lim_{x \to \infty} \frac{\sin( \frac{1}{x})}{\frac{1}{\sqrt{x}} }$$
But what next? 

l'hopital method is  applicable only  for those limits which are indeterminate . this limit can  easily  be calculated  using elemantry algebra 

