# finding the limit of a trig function

• May 27th 2012, 11:49 AM
tmas
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... :(
• May 27th 2012, 12:06 PM
Plato
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 $\left( {\frac{{2\cos (3x)}}{3}} \right)\left( {\frac{{3x}}{{\sin (3x)}}} \right)$