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? 
 A: Your idea of rewriting it that way is good when you have limit like 
$$ \lim_{x \to \infty} x^2\sin\bigg( \frac{1}{x} \bigg) $$
because you can then make the substitution $u = \frac{1}{x}$ 
However, for this problem, you should write it first as
$$ \lim_{x \to \infty} \sqrt{x}\sin\bigg( \frac{1}{x} \bigg) = \lim_{u \to \infty} u\sin\bigg( \frac{1}{u^2} \bigg) $$
by making the substitution $u = \sqrt{x}$ . Then now you have 
$$ \lim_{u \to \infty} u\sin\bigg( \frac{1}{u^2} \bigg) = \lim_{u \to \infty} \frac{ \sin\bigg( \frac{1}{u^2} \bigg)}{\frac{1}{u}} =^{L'H} \lim_{u \to \infty} \frac{\frac{-2\cos(\frac{1}{u^2})}{u^3} }{\frac{-1}{u^2} } = \frac{2\cos\bigg( \lim_{u \to \infty} \frac{1}{u^2}\bigg) }{\lim_{u \to \infty} (u) } = \frac{2}{\infty} = 0 $$
A: Consider the Taylor series expansion for $\sin(x)$. 
$$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
For $x^{-1}$, this series is
$$\frac{1}{x} - \frac{1}{3! *x^3} + \frac{1}{5! * x^5} - \ldots$$
This is equivalent to $O\Big(\frac{1}{x}\Big)$ (meaning $\sin(\frac{1}{x}) \to \frac{1}{x}$ as $x \to \infty$). So rewrite your equation as
$$\lim_{x \to \infty} \sqrt{x} *O\Big(\frac{1}{x}\Big) = O\Big(\frac{1}{\sqrt{x}}\Big)$$
Which goes to $0$ as x goes to infinity.
A: l'hopital method is  applicable only  for those limits which are indeterminate . this limit can  easily  be calculated  using elemantry algebra 

