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... :(
Re: finding the limit of a trig function
Quote:
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 }}{3}} \right)\left( {\frac{{3x}}{{\sin (3x)}}} \right))
