Squeeze Theorem

**Link88** · January 28th 2009, 07:54 PM

Use the squeeze theorem to show that lim as n goes to infinity arctan(n^2)/Sqrt(n) = 0

**Jhevon** · January 28th 2009, 08:11 PM

Originally Posted by Link88

Use the squeeze theorem to show that lim as n goes to infinity arctan(n^2)/Sqrt(n) = 0

note that $- \frac {\pi}2 \le \arctan x \le \frac {\pi}2$ for all $x$ , thus we have

$- \frac {\pi}{2 \sqrt{n}} \le \frac {\arctan (n^2)}{\sqrt{n}} \le \frac {\pi}{2 \sqrt{n}}$

can you finish?

(also, note that we can make the left side zero, but this looks nice for symmetry

)

**Link88** · January 28th 2009, 08:16 PM

Im just kinda confused on how you end up with zero from that : /

**Jhevon** · January 28th 2009, 08:39 PM

Originally Posted by Link88

Im just kinda confused on how you end up with zero from that : /

the next step is to take the limit of the system...

$\lim_{n \to \infty} - \frac {\pi}{2 \sqrt{n}} \le \lim_{n \to \infty} \frac {\arctan (n^2)}{\sqrt{n}} \le \lim_{n \to \infty} \frac {\pi}{2 \sqrt{n}}$

which means?

**Link88** · January 28th 2009, 09:39 PM

k gotcha i see how it goes to zero

Thread: Squeeze Theorem

LinkBack

Thread Tools

Display

Squeeze Theorem

Similar Math Help Forum Discussions

Squeeze Theorem

Squeeze theorem

Need help with example of Squeeze Theorem

squeeze theorem

help with squeeze theorem.

Search Tags