Question

Find the limit, if it exist.

Answer 1

\[\lim_{x \rightarrow 0}x \cos \frac{ 1 }{ x }\]

Answer 2

without using l'hopital

Answer 3

Do you know about the squeeze theorem?

Answer 4

I can't really recall it that well.

Answer 5

Basically, you use the fact that a known function is bounded and show that this other function is bounded.

Firstly, we have that \(-1\le\cos\left(\frac{1}{x}\right)\le1\), right?

Well, multiplying both sides by \(x\) gives you
\[-x\le x\cos\left(\frac{1}{x}\right)\le x\]
Taking the limit, we have
\[\lim_{x\to0}(-x)\le\lim_{x\to0}x\cos\left(\frac{1}{x}\right)\le\lim_{x\to0}x\\
0\le\lim_{x\to0}x\cos\left(\frac{1}{x}\right)\le0\]
What does this tell you about the limit?

Answer 6

That the limit equals 0?

Answer 7

Well thank you for the help Siths. Assuming what I said is the right answer from interpreting your explanation.

Answer 8

Yes, it's 0. One way of thinking about it is, even though the cosine part rapidly oscillates as it approaches 0, the factor of x "overpowers" it.