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 $\displaystyle \left( {\frac{{2\cos (3x)}}{3}} \right)\left( {\frac{{3x}}{{\sin (3x)}}} \right)$