# Proof of Limit

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• Jun 9th 2006, 11:48 AM
c_323_h
Proof of Limit
I have no experience in proofs whatsoever. Are there any general steps to go about proving math problems? I really don't know how to start these things or what the next steps are after the first step. Here is one from my calculus class I don't know how to do.

Prove
$\displaystyle \lim_{n\rightarrow 0^+}\sqrt{x} [1+sin^2(2\pi/x)]=0$
• Jun 9th 2006, 12:11 PM
CaptainBlack
Quote:

Originally Posted by c_323_h
I have no experience in proofs whatsoever. Are there any general steps to go about proving math problems? I really don't know how to start these things or what the next steps are after the first step. Here is one from my calculus class I don't know how to do.

Prove
$\displaystyle \lim_{n\rightarrow 0^+}\sqrt{x} [1+sin^2(2\pi/x)]=0$

I don't know how formal a proof you want, here is a relativle informal
proof:

As for all $\displaystyle t \in \mathbb{R}$ $\displaystyle 0 \le |\sin(t)| \le 1$ we have that:

$\displaystyle 1 \le 1+\sin^2(2\pi/x) \le 2$

So for all $\displaystyle x \ge 0$:

$\displaystyle \sqrt{x} \le \sqrt{x} [1+sin^2(2\pi/x)] \le 2\sqrt{x}$.

Now since $\displaystyle \sqrt{x} \to 0$ as $\displaystyle x \to 0$ from above this tell us that $\displaystyle \sqrt{x} [1+sin^2(2\pi/x)]$
is trapped between two function of $\displaystyle x$ both of which go to $\displaystyle 0$ as $\displaystyle x$ goes to
$\displaystyle 0$ from above and hence its limit is also $\displaystyle 0$.

RonL