Can anyone guide me for this? thank you in advance

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Let f be the function:
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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Since at each x one of the deltas work then the smaller works for all.
 
mathman said:
Since at each x one of the deltas work then the smaller works for all.
But how to find that c such that sin(c)=cos(c)?
 
Hint: sin(c) = tan(c)cos(c), and tan is surjective.

Alternatively, consider that the graphs of sin and cos intersect.

In the further alternative (not really necessary since either of the above two are sufficient to conclude that such a c exists and you don't actually need to find an exact value for it), consider the triangle formed by cutting a unit square along a diagonal.

Really, between the hint provided in the question and mathman's response you now have a complete answer to the analysis part of the question. The incidental trigonometry should have been obvious; you should be thoroughly familiar with the basic properties (geometric definitions, graphs, special values) of the trigonometric functions before studying calculus or real analysis.
 
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Equivalently, sin(c)= cos(c) is the same as saying tan(c)= 1. You should be able to solve that!
 
HallsofIvy said:
Equivalently, sin(c)= cos(c) is the same as saying tan(c)= 1. You should be able to solve that!
Erm, what i don understand is that why i have to find such c? n why sin x and cos x will be continuous at that c?eg 45 degree?
 
You will understand that when you do the final step of the proof that was sketched in the problem statement.
 
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