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I was doing a ratio test for convergence and the final expression I got before applying limit to infinity was: $\dfrac{(2+\cos(x) )}{\sqrt{x}}$, now I believe that this goes to zero, the $\dfrac{2}{\sqrt{x}}$ is trivially zero, but the $\dfrac{\cos(x)}{\sqrt{x}}$ I am having trouble trying to show it well. I know that $\cos(x)$ is bounded to finite values and the root function below is a monotonically increasing function so hence the limit should go to zero. I was wanting a better computational way to show this. Is there a better way, than what I have stated? Please let me know.

Sincerely,
Palu

Dan Rust
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Palu
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  • If you consider the part involving $\frac{2}{\sqrt{x}}$ trivial, then note that $0 \lt \frac{2+\cos x}{\sqrt{x}}\le \frac{3}{\sqrt{x}}$, and reach your conclusion by Squeezing. – André Nicolas Jun 28 '13 at 16:25

2 Answers2

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Make use of the fact that $$0 \lt \frac{1}{\sqrt x} \leq \frac{2+\cos x}{\sqrt{x}}\leq \frac{3}{\sqrt{x}}$$

and apply the squeeze theorem.

amWhy
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HINT:

As $-1\le \cos x\le 1$ for real $x$ $\implies 1\le (2+\cos x)\le3$

lab bhattacharjee
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  • Yes, I know I can do that but then I hence setup an upper bound, and then one over root(x) would go to zero. I was thinking of another way without bounding the function possibly. OK, so looks like this is the only way. Thanks. Palu – Palu Jun 28 '13 at 16:23
  • ok, so i see a sequeeze theorem approach, i forgot about that, that is good. Now I can see that it must approach zero from upper and lower. SO I wanted to ensure I can see that it must approach zero. Thanks this is what I was wanting. Really appreciate it!!! – Palu Jun 28 '13 at 16:26
  • @Palu, there may be other method(s) like http://math.stackexchange.com/questions/372681/finding-lim-x-to-infty-left1-frac-cos-x2-sqrtx-right – lab bhattacharjee Jun 28 '13 at 16:29
  • ok, thanks. Appreciate the extra materials. P – Palu Jun 28 '13 at 16:37
  • @Palu, my pleasure – lab bhattacharjee Jun 28 '13 at 18:12