# Finding limits...

• May 15th 2009, 08:54 PM
Benno_11
Finding limits...
Hi, I need some help finding the limit of the following....

$\lim_{n \to \infty} \frac{2^n - n}{3^n - n}$

There is also one more i am stuck on,

$\lim_{n \to \infty} \sqrt[n]{2^n + 3^n + 4^n}$

For this one could I use the squeeze theorem?

Thanks
• May 15th 2009, 09:40 PM
myrerer
$\lim \frac{2^n-n}{3^n-n} = \lim (\frac{2^n}{3^n-n} - \frac{n}{3^n-n}) = \lim (\frac{2^n}{3^n-n} - \frac{1}{3^n/n-1}) = \lim \frac{2^n}{3^n-n}$
$= \lim \frac{2^n}{3^n}\,\frac{1}{1-n/3^n} = \lim \frac{2^n}{3^n} = 0$.
Typically, we would just note that the linear terms ( $n$) are dominated by the higher-order terms (the exponentials), so they will drop out in the limit. (edit: what Putnam120 said)

Quote:

For this one could I use the squeeze theorem
Yes.
$4 = \sqrt[n]{0+0+4^n} < \sqrt[n]{2^n+3^n+4^n} < \sqrt[n]{4^n+4^n+4^n} < \sqrt[n]3\ 4$

In general $\lim_{n\to\infty}(\sum_i |a_i|^n)^{1/n} = \max_i |a_i|$.
• May 15th 2009, 09:50 PM
putnam120
For the first limit, when $n$ is large you have that $a^n-n$ will be almost equal to $a^n$, for $a>1$. Thus your limit is now $\lim_{n\to\infty}\left(\frac{2}{3}\right)^n=0$.