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

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

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

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

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

klimenkov
 one year ago
Best ResponseYou've already chosen the best response.1Think 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!

richyw
 one year ago
Best ResponseYou've already chosen the best response.0ah, 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?
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