Limit Question

[nietzsche]

1. The problem statement, all variables and given/known data

Find the following limit:


\lim_{x \to 0} (1-\text{cos }x)\text{sin }\frac{1}{x}


2. Relevant equations



3. The attempt at a solution

(1-cos x) -> 0 as x -> 0. sin (1/x) oscillates infinitely many times as x -> 0.

intuition tells me that the limit is 0, but how do i show that?

some ideas i have are using the fact that |sin(1/x)| =< 1, but i'm not sure.

[zcd]

Try the squeeze theorem with something that converges to zero like \frac{1-\cos{x}}{x}.

[nietzsche]

i ended up doing this.


\begin{align*}
-1 &\leq& \text{sin }\frac{1}{x} &\leq& 1\\
-(1-\text{cos }x) &\leq& (1-\text{cos }x)(\text{sin }\frac{1}{x}) &\leq& 1- \text{cos }x
\end{align*}


since both of the terms on the side equal 0 at x=0, by the squeeze theorem, the middle term also goes to 0.

[Dick]

That's how I would have done it. Well done!