Why does $\lim_{x \to 0} |{\sin (x)}|\cos (1/x) = 0$

Question

The answer $0$ was obtained by plugging it into Wolfram. I'm kind of split on the answer.

On one hand, the answer Wolfram gives makes sense because the limit of $|{\sin x}|$ evaluates to $0$. Anything multiplied by $0$ is $0$, hence why the entire limit is $0$.

But the caveat here is that the $\cos (1/x)$ part of the limit is DNE. As far as I'm aware, the rule of multiplying any number by $0$ to get a result of $0$ doesn't apply because the result of it isn't a number!

I thought about using the squeeze theorem to prove it, but I'm not sure if it works.

The key is note that $\lim\limits_{x\to 0}|\sin(x)|=0$ and $\cos(1/x)$ is a bounded function over $\mathbb{R}\setminus{0 }$. What is the result of calculating the limit of a bounded function times a function that thends to zero? — Carlos Jiménez, Sep 17 '21 at 17:09
See The problem is $\lim_{x\to 0} \sin(x)\cos\left(\frac{1}{x}\right)$ and Proving that $\lim\limits_{x\to 0} \sin(x)\cos(\tfrac1x) = 0$. — Vepir, Sep 17 '21 at 17:10
When you calculate the limit of $f(x)$ as $x \to a$, you only consider values of $x$ near to but not equal to $a$. Look carefully at the formal definition of a limit to see this. The limit can exist even if $f(a)$ is undefined. — Rob Arthan, Sep 17 '21 at 20:20

score 4 · Accepted Answer · answered Sep 17 '21 at 17:18

4

As $x\rightarrow 0$, the outputs of $\cos(1/x)$ will oscillate wildly between $-1$ and $1$. But when multiplied by $\lvert \sin(x)\rvert$ that oscillation will happen at smaller and smaller amplitudes since $\sin(x)$ will tend to zero.

answered Sep 17 '21 at 17:18

Pirate Prentice

426

score 1 · Answer 2 · answered Sep 17 '21 at 17:11

1

Hint: $\cos(\frac{1}{x}) \le 1$. Therefore

$|\sin(x)\cos(1/x)|\le |\sin(x)|$

answered Sep 17 '21 at 17:11

Richard Jensen

1,440

CiaPan · Answer 3 · 2021-09-30T08:03:20.960

0

Be careful! 'Anything multiplied by zero is zero' is certainly true, but the rule does not hold in limits. One term convergent to zero is not the same as one term equal zero. As a counterexample consider $\lim\limits_{x\to 0} \left(x\cdot\frac 1x\right)$ which equals $1$ despite the first term $x\to 0$.

edited Sep 30 '21 at 08:03

answered Sep 17 '21 at 17:42

CiaPan

13,049

Why does $\lim_{x \to 0} |{\sin (x)}|\cos (1/x) = 0$

3 Answers3