Proving Infinity of a Sequence: a(n)=[n+7]/[2+sin(n)]

[kmeado07]

Homework Statement

Prove that the following sequence (a(n)) has the property that a(n) tends to infinity as n tends to infinity.

Homework Equations

a(n)=[n+7]/[2+sin(n)]

The Attempt at a Solution

i tried l'hopitals rule, so i got 1/cos(n)...which wouldn't work.
so I am not sure how to do this question.

 

[Mark44]

L'Hopital's Rule doesn't apply here. The quotient has to be of the form 0/0 or infinity/infinity for L'H's Rule to be used.

Instead, use the fact that 1 <= 2 + sin(n) <= 3, for all n to get lower and upper limits on the value of (n + 7)/(2 + sin(n)), and then show that the lower and upper limit expressions both have the same limit as n gets large, thereby trapping the expression in the middle.

 

[Dick]

Since n+7 is unbounded, to apply l'Hopital to this you would need that the denominator is also unbounded (so you have an infinity/infinity form). It's not. The denominator is bounded and positive. That should tell you the limit right there.