lim x→0 (cos x)/x using l'hopital's

[shrewd bronze]

Online the answer says its undefined but if i use lhopitals i get, lim x→0 (-sin x)/1 = -sin(0)/1 = 0/1 = 0. Why is my answer wrong?

[surreal garnet]

This is not a limit you can apply L'hopital's rule to

[surreal garnet]

Recall the premise of L'hopital's rule

[flat grail]

l'hopital can be used in indeterminate forms such as inf/inf or 0/0. This is 1/0 which is undefined

[shrewd bronze]

ahhh okay, just one more thing however, in my uni answers they explained by using x = 1/n but im not sure how thats supposed to work

Basically: For the sequence xn = 1/n is limn→∞ xn = 0 and since cos is monotonally decreasing on [0, 1], it holds for all n ∈ N

[flat grail]

I dont get this

why is there cos(1)

[surreal garnet]

Same, why not just cos(0)

But you can pick pretty much any value of cos

[flat grail]

true but why the hell cos(1)? What makes it so special

[surreal garnet]

There is a theorem saying that the limit of f(x) as x -> a exists when all of the sequences x_n such that x_n approach a as n -> infinity you still have the same limit

Basically, if you can come up with a sequence such that the limit diverges

Then the original limit doesn't exist

This is what they have done here

They took x_n = 1/n (because x_n approaches 0 as n -> infinity) and showed that cos(x_n)/x_n >= ncos(1)

But, as we already know, ncos(1) diverges

Thus the limit of cos(x)/x as x -> 0 doesn't exist

[shrewd bronze]

why did they choose cos(1) here though?

[surreal garnet]

We have no idea as well

[flat grail]

I guess that they actually wanted to use cos(0), probably typo

[surreal garnet]

But you could pick any value of cos

[shrewd bronze]

yeah could be, our uni has a bunch of those in theri answers tbh

[surreal garnet]

Would be weird to make the same typo twice

[shrewd bronze]

true

[surreal garnet]

But, nonetheless, the reasoning is correct

[shrewd bronze]

alright, thanks a lot!!

[surreal garnet]

Similar argument (as in involving negating the theorem mentioned) can be used to prove that any non-constant periodic function doesn't have a horizontal asymptote btw

[shrewd bronze]

negating the theorem?

[surreal garnet]

You don't know what it means to negate a statement?

[shrewd bronze]

well the opposite, but im not sure why you would need to negate the theorem, when it already shows that some non-constant periodic functions dont have a horizontal asymptote as in this case with cosx/x

[surreal garnet]

1. cosx/x is not a periodic function
2. The theorem is used to prove existence of some limits, you can't use it to disprove limits until you negate it or show a contradiction

[strong timber]

Also because it would be a circular argument, since this limit is the derivative of cos(x) at 0

The same reason why you can't use L'Hopital's for sin(x)/x as x -> 0

[terse isle]

.solved