sequence, find the limit.

**rcmango** · January 30th 2007, 06:42 PM

heres the equation for this sequence: http://img401.imageshack.us/img401/6712/untitledfe2.jpg

i'm asked to find the limit. I thought i could use the dqueeze theorem, i may have been on track using 1/sqrt(n) ?

help need please.

**ThePerfectHacker** · January 30th 2007, 06:51 PM

Note that,
$-1\leq \sin x \leq 1$
Adding 1,
$0\leq 1+\sin x\leq 2$
Divide by $1+\sqrt{n}>0$
Thus,
$0=\frac{0}{1+\sqrt{n}}\leq \frac{1+\sin x}{1+\sqrt{n}}\leq \frac{2}{1+\sqrt{n}}$

The limits,
$\lim_{n \to \infty} 0 =0$

$\lim_{n\to \infty} \frac{2}{1+\sqrt{n}}=0$

Thus,
$\lim_{n\to \infty} \frac{1+\sin x}{1+\sqrt{n}}=0$

**rcmango** · January 30th 2007, 08:51 PM

i understand. thanks.

also, where you added one. you state +sin because u also added +1 to the middle term also, correct?

thanks alot, excellent.

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