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So I have a problem where I am to determine the limit or prove it doesn't exist.

The problem is:

$$ \lim_{x\to 0} \sin(x)\cos\left(\frac{1}{x}\right) $$

My proof is as follows (I'm pretty confident about this):

For all $ x \ne 0, -1 \le \cos\left(\frac{1}{x}\right) \le 1$.

Hence, $-\sin(x) \le \sin(x)\cos\left(\frac{1}{x}\right) \le \sin(x) $

As $x$ goes to zero, so does $-\sin(x)$ and $\sin(x)$.

So by squeeze lemma, the original problem has a limit of $0$.

So my question is, is this sufficient enough where everything seems trivial? The main part I'm worried about is the $\sin(x) = 0$ as $x$ goes to $0$. Do I need to prove this? I just assumed it was ok because $\sin(0) = 0$.

Mark Viola
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2 Answers2

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You're almost correct. Instead of writing things as in the OP, we can simply write

$$0\le |\sin(x)\cos(1/x)|=|\sin(x)|\,|\cos(1/x)|\le |\sin(x)|\,(1)=|\sin(x)|$$

And now apply the squeeze theorem.

And if there are any doubts on the limit of the sine function, recall that we have

$$|\sin(x)|\le |x|$$

for all $x$.

Mark Viola
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Your inequalities need to be flipped based on what half of the plane we are in. We have that $$-\sin(x) < \sin(x)\cos(1/x) < \sin(x) \qquad \forall x> 0$$ $$\sin(x) < \sin(x)\cos(1/x) < -\sin(x) \qquad \forall x< 0$$ Can you see why this is? Now, this doesn't really affect your squeezing argument, so you are still fine.

Brevan Ellefsen
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