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richywBest ResponseYou've already chosen the best response.0
\[f(x)=x\sin{\left(\frac{1}{x}\right)} \text{ if } x\neq0\]\[f(0)=0\]
 one year ago

richywBest ResponseYou've already chosen the best response.0
my textbook shows that\[\lim_{x\rightarrow 0}x\sin{\left(\frac{1}{x}\right)=0}\]because\[\leftx\sin{\left(\frac{1}{x}\right)}\right\leqx\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
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 \(x=0\).
 one year ago

richywBest ResponseYou've already chosen the best response.0
I don't understand how this shows that the limit is zero?
 one year ago

richywBest ResponseYou've already chosen the best response.0
I mean it's clear that this inequality holds to me. but it's unclear to me how this finds the limit.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.1
Think about it. If \(\leftx\sin{\left(\frac{1}{x}\right)}\right\leqx\) and \(x\rightarrow0\), so \(x\) is very little, and then \(\leftx\sin{\left(\frac{1}{x}\right)}\right \) is little too!
 one year ago

richywBest ResponseYou've already chosen the best response.0
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?
 one year ago
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