What is the limit value of this function (sin..)

Question

I wanted to ask you a question about this specific function:

$$\lim\limits_{x\to\infty}\left( \sin(\sqrt{x})-\sin(\sqrt{x+1})\right)$$

Somehow I can't comprehend how to do this task On one hand sin has its specific attributes like jumping around $1$ and $- 1$.. So it's divergent ? Or does it converge against $0$.. I don't really know It would be nice if you could give me a hint/solution

score 2 · Answer 1 · answered Jan 20 '15 at 15:52

2

Hint: Use the inequality $$|\sin x-\sin y|\le |x-y|, \forall x,y$$ and then apply the squeeze theorem.

answered Jan 20 '15 at 15:52

Tien Kha Pham

2,980

Quite a splendid solution!! – Vim Jan 20 '15 at 16:02
Aren't you missing - x - y <= sin x - y ? – Kusottare Jan 21 '15 at 15:54
@Kusottare I think you are implying the inequality $|\sin x|\le x$, but in my opinion it makes no sense here. – Tien Kha Pham Jan 21 '15 at 16:42
Ok I will answer you with a new comment about my ideas.. could you correct improve them ? I think it makes more sense this way for me – Kusottare Jan 21 '15 at 20:33

score 0 · Answer 2 · answered Jan 21 '15 at 20:41

Because x+1> x I can change the algebraic signs in sin (sin(sqrt(x) - sin(sqrt(x+1)) = -2 lim x -> infinity cos (sqrt(x) + sqrt(x+1)/2) sin (sqrt(x+1 - x )/2)

Now I expand my sin (sin(sqrt(x) - sin(sqrt(x+1)) = -2 lim x -> infinity cos (sqrt(x) + sqrt(x+1)/2) sin ((sqrt(x+1) - (sqrt(x+1))

(sin(sqrt(x) - sin(sqrt(x+1)) = -2lim -> infinity cos (sqrt(x) + sqrt(x+1))/2 sin (1/2(sqrt(x+1)+sqrt(x))

sin goes against 0 because the denominator goes against 0

What is the limit value of this function (sin..)

2 Answers2