Here's the question you clicked on:
richyw
show that f is continuous at every point, \(a\in\mathbb{R}\)
\[f(x)=x\sin{\left(\frac{1}{x}\right)} \text{ if } x\neq0\]\[f(0)=0\]
my textbook shows that\[\lim_{x\rightarrow 0}x\sin{\left(\frac{1}{x}\right)=0}\]because\[\left|x\sin{\left(\frac{1}{x}\right)}\right|\leq|x|\]
The function is continious in point \(x_0\) if \(\lim \limits_{x\rightarrow x_0}f(x)=f(x_0)\). You need to show that the function is continious in the point \(x=0\).
I don't understand how this shows that the limit is zero?
I mean it's clear that this inequality holds to me. but it's unclear to me how this finds the limit.
Think about it. If \(\left|x\sin{\left(\frac{1}{x}\right)}\right|\leq|x|\) and \(x\rightarrow0\), so \(|x|\) is very little, and then \(\left|x\sin{\left(\frac{1}{x}\right)}\right| \) is little too!
ah, perfect. I get it now. so now because the limits exist, and f(0) is defined, and the limit= f(0), that should be satisfactory to show that the function is cts right?