Find the limit of the following: $\lim_{n \to \infty}(n^2+1)(\cos(\frac{1}{n})-1)$ $$\lim_{n \to \infty}(n^2+1)\left(\cos\left(\dfrac{1}{n}\right)-1\right)$$  
Now I have been working on this for a while but don't know how to proceed, using L'Hopital isn't helping me, no matter which term I take the reciprocal of and divide by, I still get $0/0$ limits as I differentiate multiple times.
Any help would be appreciated.
 A: Using the trigonometric formula
$$1-\cos(x)=2\sin^2\left(\frac{x}{2}\right)$$ gives us
$$\cos\left(\frac{1}{n}\right)-1=-2\sin^2\left(\frac{1}{2n}\right)$$
Now
$$L=\lim_{n \to \infty}(n^2+1)\left(\cos\left(\frac{1}{n}\right)-1\right)=-2\lim_{n \to \infty}(n^2+1)\cdot\sin^2\left(\frac{1}{2n}\right)$$
Therefore$$L=-2\lim_{n \to \infty}\frac{1}{2n}\cdot\frac{1}{2n}\cdot(n^2+1)\cdot\frac{\sin\left(\frac{1}{2n}\right)}{\frac{1}{2n}}\cdot\frac{\sin\left(\frac{1}{2n}\right)}{\frac{1}{2n}}$$
The two fractions involving sines tend to $1$. So the limit is $-\frac{1}{2}$
A: The ${}+1$ adds nothing to the limit. That is,
$$\lim_{n\to\infty}(n^2+1)\left(\cos\left(\frac1n\right)-1\right)=\lim_{n\to\infty}n^2\left(\cos\left(\frac1n\right)-1\right)$$
since $\lim_{n\to\infty}\left(\cos\left(\frac1n\right)-1\right)=0$.
Now you have $$\begin{align}\lim_{n\to\infty}n^2\left(\cos\left(\frac1n\right)-1\right)
&=\lim_{n\to\infty}\frac{\cos\left(\frac1n\right)-1}{\frac{1}{n^2}}\\
&=\lim_{x\to0}\frac{\cos(x)-1}{x^2}\end{align}$$
which is well-known to be ${-{\frac{1}{2}}}$, either by L'Hospital's Rule, Taylor Series, or by a squeeze theorem argument that is also used to prove the slightly more well-known limit $\lim_{x\to0}\frac{\sin(x)}{x}=0$.
A: It might help to break up the limit after the application of L'Hospital:
$$
\begin{align}
\lim_{n\to\infty}\frac{\cos\left(\frac1n\right)-1}{\frac1{n^2+1}}
&=\lim_{n\to\infty}\frac{\frac1{n^2}\sin\left(\frac1n\right)}{-\frac{2n}{(n^2+1)^2}}\\
&=-\frac12\lim_{n\to\infty}\left(\frac{n^2+1}{n^2}\right)^2\lim_{n\to\infty}\frac{\sin\left(\frac1n\right)}{\frac1n}
\end{align}
$$
