Find and prove
$\displaystyle \lim_{n\rightarrow\infty}(\sqrt{n+3} - \sqrt{n})$
From
$\displaystyle a_n=\frac{3}{\sqrt{n+3}+\sqrt{n}}$
You can show that this is always less than
$\displaystyle b_n=\frac{3}{\sqrt{n}+\sqrt{n}}$
and always greater than
$\displaystyle c_n=0$
Using the sandwich theorem show that a_{n} converges to zero.
right but when proving, for all epsilon greater than 0, what N would I choose so that $\displaystyle |\frac{3}{2\sqrt{n}}| < \epsilon$
If i try to square it, epsilon gets squared too...
Right so we have this:
$\displaystyle |\frac{3}{\sqrt{n} + \sqrt{n}}| < \epsilon$
$\displaystyle \Rightarrow |\frac{3}{2\sqrt{n}}| < \epsilon$
$\displaystyle \Rightarrow \frac{3}{2\sqrt{n}} < \epsilon$
Now from this point on, if I try to isolate n, at some point I will have to square $\displaystyle \sqrt{n}$ which will yield $\displaystyle {\epsilon}^2$ ...