# Proof of Limit

• Jun 9th 2006, 12:48 PM
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
$\lim_{n\rightarrow 0^+}\sqrt{x} [1+sin^2(2\pi/x)]=0$
• Jun 9th 2006, 01:11 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
$\lim_{n\rightarrow 0^+}\sqrt{x} [1+sin^2(2\pi/x)]=0$

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

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

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

So for all $x \ge 0$:

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

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

RonL