Thread: finding the limit of a trig function

I know i got this one wrong somewhere along the way...

lim x goes to 0 (2x)cot(3x)

this is what I did:

lim x goes to 0 ((2xcos3x)/sin3x) = lim x to 0 (2xcos3x)/3x time lim to 0 sin 3x/3x
= 2 times lim x to 0 cos3x/3x times lim x to 0 sin 3x/3x
both cose x/x and sinx/x cancel out to one
therefore: 2(3)(3) = 18

I think I messed up on the manipulation somewhere...

Originally Posted by tmas

I know i got this one wrong somewhere along the way...

lim x goes to 0 (2x)cot(3x)

Write it as