What is the limit of this function as $n$ tends to infinity?

Question

$\lim_{n\rightarrow\infty}\sqrt{n}\cos(\frac{\pi}{2}-\frac{1}{\sqrt{n}})$

I'm having a lot of trouble figuring it out. My first step is always to convert $\cos(\frac{\pi}{2}-\frac{1}{\sqrt{n}})$ to $\sin(-\frac{1}{\sqrt{n}})$ and then I get stuck here. Because I'm not quite sure where $\lim_{n\rightarrow\infty}\sqrt{n}(-\sin(\frac{1}{\sqrt{n}})$ leads....:/

Please help.

EDIT

My Thomas' Calculus text book (12th Edition) lists the identity as being $$cos(A-\frac{\pi}{2}) = sin(A)$$ so naturally (or perhaps, naively?) I went ahead and took my A to be $-\frac{1}{\sqrt{n}}$

Answer 1

$\cos(\frac\pi2-x)=\sin x$

$\cos(\frac\pi2-\frac 1{\sqrt n})=\sin \frac 1{\sqrt n}$

Put $h=\frac 1{\sqrt n},$ so, $h\to 0$ as $n\to ∞$

So, $\lim_{n\rightarrow\infty}\sqrt{n}\cos(\frac{\pi}{2}-\frac{1}{\sqrt{n}})$

$=\lim_{n\rightarrow\infty}\sqrt{n}\sin(\frac 1{\sqrt{n}})$

$=\lim_{ h\to 0}\frac{\sin h}{h}=1$

Answer 2

Hint

$\lim\limits_{n\rightarrow\infty}\sqrt{n}\cos\left(\dfrac{\pi}{2}-\dfrac{1}{\sqrt{n}}\right)=\lim\limits_{n\rightarrow\infty}\sqrt{n}\sin\left(\dfrac{1}{\sqrt{n}}\right)=\lim\limits_{n\rightarrow\infty}\dfrac{\sin\left(\dfrac{1}{\sqrt{n}}\right)}{\dfrac{1}{\sqrt{n}}}.$ What did you know about $\lim\limits_{x\rightarrow 0}\dfrac{\sin{x}}{x}$?