2004-06-30
Hello, Dave.
I finally came up with a solution to the problem, thanks to you
If you use the definition for continuity (like you said), combined with the subtraction theorum for sine and the fact that sin(u)<=|u| and
cos(u)<=1 then what you get is this: (here "<=" means "smaller than or equal to").
0<=|sin(x) - sin(a)| = (subtraction formula for sine)
= 2|sin(x-a)/2|*|cos(x+a)/2|<= (cos(u)<=1, sin(u)<=|u|) <= 2|(x-a)/2|
<=|x-a|
This means that the function |sin(x) - sin(a)| is "trapped" between 0 and the function |x - a|, which goes to 0 as x - >a.
Hence:
lim |sin(x) - sin(a)| = 0
x->a
which is the same as if:
lim sin(x) = sin(a)
x->a
For all x.
Q.E.D
Thanks for the hint
Hans Lindroth.