Limit (x->0) (Cos 2x) / 3x = ?

Question

(Calc 1): Been struggling with this one - I'll bet I'm missing something easy... Any guidance would be appreciated.

Answer 1

As x -> 0+, 3x -> 0+ and cos(2x) -> 1. Since cos is an even function, it doesn't matter if x ->0+ or 0-, we always gert 1.

So, as x -> 0+ cos(2x)/(3x) -> oo, there's no finite limit.

And as x->0- cos(2x)/(3x) -> -oo

Answer 2

If you try to evaluate the expression at x = 0 you get 1/0, which is undefined. You can't use L'hopital's rule since the expression does not give you 0/0 or infinity / infinity.
Consider the following argument:
cos(2x) is bounded between -1 and 1, but 3x is unbounded. Thus as the bottom approaches zero, the quotient grows without bound. Thus there is no limit.
If x was going to infinity, the same argument would show you that the limit was 0.

Answer 3

you're not missing anything simple. it has no limit.
cos(0) is 1
3(0-) = 0- (approaching 0 from left)
3(0+)= 0+ (approaching 0 from right)

One is appoaching -infinity, the other +infinity. So no limit exists.

Comment on Harry M's answer: the cos2x/2x trick works only if you have sin(2x)/2x

Answer 4

Lim cos (2x) / (3x) is infinity as x approaches 0!

Answer 5

well cos 0=1 so 1/0=infinte, well dude i think u missed the cosin(0)=1

Answer 6

Limit (x->0) (Cos 2x) / 3x
=Limit (x->0) (Cos 2x)*2 / 2*3x
=Limit (x->0) (Cos 2x) / 2x*2/3
= 1*2/3
=2/3