How can I use the sequential criterion for functional limits to show that the following limit exist and compute the limit:
$$\lim_{x \rightarrow 0} \sqrt{|x|}\cos\left(1/x\right) \ \text{for} \ x \in \mathbb{R} \backslash \{0\}$$
My attempt:
I have a hunch the limit is 0, but I can't really use the sequential criterion to show it is 0. What I've done so far is just try one particular sequence, i.e., define $x_n = \frac{1}{2 n \pi}$, clearly $(x_n) \rightarrow 0$.
Let $f(x) = \sqrt{|x|}\cos\left(1/x\right)$. Then $f(x_n) = \sqrt{\frac{1}{2 n \pi}}\cos\left(2n \pi\right) = \sqrt{\frac{1}{2 n \pi}}$. So $f(x_n) \rightarrow 0$.
Now the thing is, this is only one particular sequence, to use the sequential criterion, I need to show for ALL sequences.
How do I finish the question?
Working off the hint:
Let $(x_n)$ be any sequence in $\mathbb{R} \backslash \{0\}$ such that $(x_n) \rightarrow 0$. Then $0 \le \sqrt{|x_n|}\cos\left(1/x_n\right) \le \sqrt{|x_n|}$, using the squeeze theorem, we have $f(x) \rightarrow 0$.
Since $(x_n)$ was arbitrary, we are done.
Best Answer
Hint: Choose any sequence $x_n$ in $\mathbb{R} \setminus \{0\}$ converging to zero. Using the fact that $|\cos x| \le 1$ for all $x$,
$$|f(x_n)| \le \sqrt{x_n}$$