The limit of sinx/x as x approaches infinity is 0?

Question

Okay, so I understand that the answer is zero... I don't know how to show it with l'Hopital's Rule. When I applied the rule, I got the limit of cosx as x approaches infinity.. is that right so far? What would I do next?

THANKS in advance!!!

Answer 1

you can't use l'hopital's rule for this since there is no limit for sin x as x tends to infinity, hence you cant use the rule.



however you can show it using the sandwich rule (basically if your function is between two other functions with the same limit, then your function will have this limit too)



now we know that sin x is always between 1 and -1. So we see that



-1/x <= (sin x)/x <= 1/x



(where the <= is less than or equal to)



Now as x-> infinity we know that both 1/x and -1/x tend to 0, so (sin x)/x must do, by the sandwich rule.

Answer 2

Limit Approaching Infinity

Answer 3

Should be 0

Answer 4

We have limit x tends to infinity sin x/ x equal to product of limit x tend to infinity sin x and limit x tend to 0 1/x that is product of a number between -1 to 1 and 0 which is equal to 0.

Answer 5

L'Hôpital's rule is applied to indeterminate forms, both 0/0 and infinity/infinity. But this rule holds only if the limit of f'(x)/g'(x) exists.



In your case, (sin(x))'/x'=cos(x). The limit of cos(x) as x approaches infinity don't exist so we can't use the l'Hôpital's rule in this case.

Answer 6

yes that is right, but i don't think the limit of cos(x) going to infinity exists.

Answer 7

yes