The function is continuous at 1 so .
I have this problem:
lim (x-->1) sin(2x)/3x
I multiplied the whole thing by 2/2 and took a 2/3 out of the limit, so I then have:
2/3 [lim(x-->1) sin(2x)/2x
Which of course is very similar to the sinx/x limit, but the variable is approaching 1, not 0 in this case...
Would it be correct to do this:
Let z = x-1
x-->1, then z-->0
and turn the equation into:
(2/3) [lim(z-->0) sin2(z+1)/2(z+1)
I feel like this isn't going to get me anywhere useful...
Any tips, please? I appreciate it.