Use the limit rule to find $\lim_{n\to\infty} n^{2}(1-\cos(\frac{1}{n}))$ I managed to do this with L'hospitals rule but can I use trig identities to make this simpler?
 A: Multiply top and (missing) bottom by $1+\cos(1/n)$. After using the identity $1-\cos^2 t=\sin^2 t$, we get that we want
$$\lim_{n\to\infty}\left(\frac{1}{1+\cos(1/n)}\cdot \frac{\sin^2(1/n)}{(1/n)^2}\right).$$ 
The rest should follow from a limit you know. 
A: Using $$\cos x-\cos y =2\sin\frac{x+y}{2}\sin\frac{y-x}{2}$$
we find with $x=0$ and $y=\frac1n$
$$ 1-\cos\frac1n=2\sin^2\frac1{2n}$$
Now if you know $\lim_{x\to 0}\frac{\sin x}x=1$, you also get $\lim_{n\to\infty}2n\sin\frac1{2n}=1$ so ultimately
$$\lim_{n\to\infty} n^2\left(1-\cos \frac1n\right) =\lim_{n\to\infty}2n^2\sin^2\frac1n=\frac12\left(\lim_{n\to\infty}2n\sin\frac1{2n}\right)^2=\frac12.$$
A: With equivalents:
It's a classic fact that $\;1-\cos\dfrac1n\sim_\infty\dfrac1{2n^2}$ (for a proof see Taylor development at order $2$ of $\cos$), hence $$n^2\Bigl(1-\cos \frac1n\Bigr)\sim_\infty\frac{n^2}{2n^2}=\frac12.$$
A: Start with the identity $1-\cos(2x)=2\sin^2(x)$ to get
$$
\begin{align}
\color{#C00000}{n^2}\left(\color{#00A000}{1-\cos\left(\frac1n\right)}\right)
&=\frac{\color{#00A000}{2\sin^2\left(\frac1{2n}\right)}}{\color{#C00000}{\frac1{n^2}}}\\
&=\frac12\left(\frac{\sin\left(\frac1{2n}\right)}{\frac1{2n}}\right)^2\tag{1}
\end{align}
$$
Then use the limit
$$
\lim_{x\to0}\frac{\sin(x)}x=1\tag{2}
$$
to get
$$
\lim_{n\to\infty}n^2\left(1-\cos\left(\frac1n\right)\right)=\frac12\tag{3}
$$
A: $$\lim\limits_{n\to\infty} n^2\left(1-\cos\left(\frac{1}{n}\right)\right)$$
Let $m=\frac1n$, then
$$\lim\limits_{m\to 0^+} \frac{1-\cos m}{m^2}=\lim\limits_{m\to 0^+} \frac{2\sin^2\left(\frac{m}{2}\right)}{m^2}$$
Let $k=\frac{m}{2}$, then
$$\lim\limits_{k\to 0^+} \frac{2\sin^2 k}{4k^2}=\frac12\lim\limits_{k\to 0^+} \frac{\sin^2 k}{k^2}=\frac12$$
