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richyw
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show that f is continuous at every point, \(a\in\mathbb{R}\)
 one year ago
 one year ago
richyw Group Title
show that f is continuous at every point, \(a\in\mathbb{R}\)
 one year ago
 one year ago

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richyw Group TitleBest 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

richyw Group TitleBest 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

klimenkov Group TitleBest 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

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

richyw Group TitleBest 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

klimenkov Group TitleBest 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

richyw Group TitleBest 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

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
Yes. That is right.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
thank you so much
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.1
You are welcome.
 one year ago
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