Let f be the function:(adsbygoogle = window.adsbygoogle || []).push({});

f(x) =

sin(x) ; x is element of Q

cos(x) ; x is not element of Q

Prove, using epsilon-delta definition, that there is a point c,which is element of R at which f is continuous.

Hint: Consider c such that sin(c) = cos(c); why does such a c exist? Then,

since you know that sin(x) and cos(x) are continuous at c, for epsilon> 0, you get delta1 >

0 that gives lsin(x)-sin(c)l <epsilon , and also delta2 > 0 that gives lcos(x)-cos(c)l <epsilon .

Now, why does delta = min(delta1; delta2) work to show lf(x)- f(c)l < epsilon.

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# Can anyone guide me for this question?

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