How Can the Inequality -x ≤ sin(x) ≤ x Help Prove a Convergent Integral?

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SUMMARY

The discussion focuses on proving the inequality -x ≤ sin(x) ≤ x for x > 0, which is essential for demonstrating the convergence of the integral ∫^1_0 nxsin(1/nx)dx to 1. The participant successfully established the inequality by integrating both sides of cos(t) ≤ 1 between 0 and x. Additionally, they referenced the limit of the integral ∫^T_1 cos(x)/√x dx as T approaches infinity, confirming its existence, which supports their argument for the convergence of the integral in question.

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Homework Statement


I am attempting to show that [itex]-x \leq sin(x) \leq x[/itex] for x>0 and thus [itex]\int^1_0 nxsin(\frac{1}{nx})dx[/itex] converges to 1.


Homework Equations



I know that I need to use the fact that I have shown that the limit as T tends to infinity of [itex]\int^T_1 \frac{cos(x)}{\sqrt{x}}dx[/itex] exists.


The Attempt at a Solution


 
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henry22 said:

Homework Statement


I am attempting to show that [itex]-x \leq sin(x) \leq x[/itex] for x>0 and ...

The -x part is trivial and for the rest integrate both sides of cos(t) ≤ 1 between 0 and x, x > 0.
 
LCKurtz said:
The -x part is trivial and for the rest integrate both sides of cos(t) ≤ 1 between 0 and x, x > 0.

OK I've done this and I get the inequality I need. But can I just check, I don't understand how I have used the equation I need to in the OP?

For the second part if I know that -x<= sinx <= x then -1<=nx(sin(1/nx)) <= 1 but then I'm a bit stuck
 

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