# find the limit if it's converge

• Mar 20th 2010, 04:30 PM
wopashui
find the limit if it's converge
$\displaystyle 2^n/(4^n+1)$

i know it's converge to 0, but I have trouble dealing with the n expontial, can someone tell me the general methof for finding the limits for $\displaystyle a^n/(b^n + 1)$
• Mar 20th 2010, 04:53 PM
Random Variable
$\displaystyle \lim \frac{2^{n}}{1+4^{n}} =\lim \frac{\frac{2^{n}}{4^{n}}}{\frac{1}{4^{n}}+\frac{4 ^{n}}{4^{n}}}$

$\displaystyle \lim \frac{\frac{1}{2^{n}}}{\frac{1}{4^{n}} + 1} = \frac{0}{0+1} = 0$
• Mar 20th 2010, 06:03 PM
wopashui
thanks, i have got another one, $\displaystyle n^2/\sqrt{2n^4+1}$

how do I deal with the squre root?
• Mar 20th 2010, 06:10 PM
Random Variable
Quote:

Originally Posted by wopashui
thanks, i have got another one, $\displaystyle n^2/\sqrt{2n^4+1}$

how do I deal with the squre root?

Divide the numerator and the denominator by $\displaystyle n^{2}$. When you bring the $\displaystyle n^{2}$ inside of the radical it becomes $\displaystyle n^{4}$
• Mar 20th 2010, 09:33 PM
wopashui
Quote:

Originally Posted by Random Variable
Divide the numerator and the denominator by $\displaystyle n^{2}$. When you bring the $\displaystyle n^{2}$ inside of the radical it becomes $\displaystyle n^{4}$

hmm, I dun quite understand it, can you show me the steps please?