Find $\lim\limits_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$ $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)$$

I'm surprisingly struggling with this limit, could you give me a hint how to handle it (no L'Hospital and no prior knowledge what the limit value is)? 

Generally what are some basic methods to handle $\infty \cdot 0$?
Hint:

Set $\dfrac1{\sqrt n}=2h,n=\dfrac1{4h^2}$

as $n\to\infty, h\to0^+$

and use  $$\cos2h=1-2\sin^2h$$
Enforcing the change of variables  $t^2 =1/n $ then   $$\lim_{n \rightarrow \infty}n\bigg(\cos \bigg(\frac {1}{\sqrt n} \bigg) - 1\bigg)= - \lim_{t \rightarrow 0}\frac{1-\cos(t) }{t^2}=-\frac12 .$$

This last limit is classical for trigonometric function