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richyw
 3 years ago
show that f is continuous at every point, \(a\in\mathbb{R}\)
richyw
 3 years ago
show that f is continuous at every point, \(a\in\mathbb{R}\)

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

richyw
 3 years 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
 3 years 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
 3 years 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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