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.
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.
Note that,
$\displaystyle -1\leq \sin x \leq 1$
Adding 1,
$\displaystyle 0\leq 1+\sin x\leq 2$
Divide by $\displaystyle 1+\sqrt{n}>0$
Thus,
$\displaystyle 0=\frac{0}{1+\sqrt{n}}\leq \frac{1+\sin x}{1+\sqrt{n}}\leq \frac{2}{1+\sqrt{n}}$
The limits,
$\displaystyle \lim_{n \to \infty} 0 =0$
$\displaystyle \lim_{n\to \infty} \frac{2}{1+\sqrt{n}}=0$
Thus,
$\displaystyle \lim_{n\to \infty} \frac{1+\sin x}{1+\sqrt{n}}=0$