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 $\infty \cdot 0$? 
 A: Hint:
Set $\dfrac1{\sqrt n}=2h,n=\dfrac1{4h^2}$
as $n\to\infty, h\to0^+$
and use  $$\cos2h=1-2\sin^2h$$
A: $$\lim _{ n\rightarrow \infty  } n\left( \cos  \left( \frac { 1 }{ \sqrt { n }  }  \right) -1 \right) =\lim _{ n\rightarrow \infty  } n\left( \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) - }  } \cos ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) -\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right) - }  }  \right) =\\ =\lim _{ n\rightarrow \infty  } n\left( -2\sin ^{ 2 }{ \left( \frac { 1 }{ 2\sqrt { n }  }  \right)  }  \right) =-\frac { 1 }{ 2 } \lim _{ n\rightarrow \infty  }{ \left( \frac { \sin { \left( \frac { 1 }{ 2\sqrt { n }  }  \right)  }  }{ \frac { 1 }{ 2\sqrt { n }  }  }  \right)  } ^{ 2 }=-\frac { 1 }{ 2 } $$
A: $$ \cos  \left( \frac { 1 }{ \sqrt { n }  }  \right)=1-\frac{1}{2n}+o\left(\frac1n\right)$$
$$n\left( \cos \left( \frac { 1 }{ \sqrt { n }  } \right) -1 \right)=n\left( 1-\frac{1}{2n}+o\left(\frac1n\right) -1 \right)=-\frac12+o(1)\to-\frac12$$
A: Possible solution : 
$$l = \lim_{n\to\infty}\frac{e^{\ln\cos\big(\frac{1}{\sqrt{n}}\big)}-1}{\ln\cos\big(\frac{1}{\sqrt{n}}\big)}\cdot\frac{\ln\cos\big(\frac{1}{\sqrt{n}}\big)}{\frac{1}{n}} = \frac{1}{2}\lim_{n\to\infty}\frac{\ln\big(1-\sin^2(\frac{1}{\sqrt{n}})\big)}{-\sin^2(\frac{1}{\sqrt{n}})}\cdot\frac{-\sin^2(\frac{1}{\sqrt{n}})}{(\frac{1}{\sqrt{n}})^2} = -\frac{1}{2}$$
Using $$\lim_{x\to0}\frac{e^x-1}{x} = \lim_{x\to0}\frac{\ln(1+x)}{x} = \lim_{x\to0}\frac{\sin x}{x}=1$$
A: For $x\approx 0,\sin x\approx \tan x\approx x$
We let $x=\dfrac1n$, so as $n\to\infty,x\to0$
$$\begin{align}L&=\lim_\limits{x\to0}\dfrac{\cos \sqrt x-1}x\\&=\lim_\limits{x\to0}\dfrac{\sqrt{1-\sin^2\sqrt x}-1}{x}\\&=\lim_\limits{x\to0}\dfrac{\sqrt{1-x}-1}x\cdot\dfrac{\sqrt{1-x}+1}{\sqrt{1-x}+1}\\&=\lim_\limits{x\to0}\dfrac{-x}{x(\sqrt{1-x}+1)}\\&=\lim_\limits{x\to0}\dfrac{-1}{\sqrt{1-x}+1}\\&=-\dfrac12\end{align}$$
A: By the fundamental trigonometric limit you have
$$\lim_{x \to 0} \frac{\cos(x)-1}{x^2}=-\frac{1}{2}$$
It follows that if $a_n \neq 0$ is any sequence converging to $0$ we have
$$\lim_{n \to \infty} \frac{\cos(a_n)-1}{a_n^2}=-\frac{1}{2}$$
Set $a_n=\frac{1}{\sqrt{n}}$.
A: If you want to go beyond the limit itself, considering$$A=x\left(\cos \left(\frac {1}{\sqrt x} \right) - 1\right)$$ define $x=\frac 1 {t^2}$ making $$A=\frac 1 {t^2}\left(\cos \left(t \right) - 1\right)$$ Now, use the classical Taylor expansion 
$$\cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^6\right)$$ which makes
$$A=-\frac{1}{2}+\frac{t^2}{24}+O\left(t^4\right)$$ which shows the limit when $t\to 0$ and also how it is approached.
If you want to go back to $n$, replace $t$ by $\frac 1 {\sqrt n}$ to get $$n\left(\cos \left(\frac {1}{\sqrt n} \right) - 1\right)=-\frac{1}{2}+\frac{1}{24n}+O\left(\frac 1 {n^2}\right)$$
A: 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 
