Evaluating a tricky limit I'm trying to calculate the radius of convergence for a power series and I'm stuck at the following step:
$$\lim_{n\to \infty}\Bigg\lvert\left(\frac{3^n+\left(-4\right)^n}{5}\right)^{\frac{1}{n}}\Bigg\rvert$$
How would you evaluate this limit?
 A: For $n$ odd, we see that 
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}&=\dfrac{3}{5^{1/(2k+1)}}\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)},
\end{align*}
where
\begin{align*}
\lim_{k\rightarrow\infty}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)}&=\lim_{k\rightarrow\infty}\dfrac{1}{2k+1}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)\\
&=\lim_{k\rightarrow\infty}\dfrac{1}{2}\dfrac{2\cdot(4/3)^{2k+1}\log(4/3)}{(4/3)^{2k+1}-1}\\
&=\log(4/3),
\end{align*}
so 
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}\rightarrow\dfrac{3}{1}\cdot e^{\log(4/3)}=4.
\end{align*}
For $n$ even, the same procedure leads to the limit $4$.
A: Simply observe by algebraic rules for limits:
$$\left(\left|\left(\frac{3^n+\left(-4\right)^n}{5}\right)^{\frac{1}{n}}\right|\right)=\\=|-4|\cdot \left(\left|\left(\frac{{\frac{-3}{4}}^n+\left(1\right)^n}{5}\right)^{\frac{1}{n}}\right|\right) \to \\ \to 4\cdot 1=4$$
Indeed note that:
$$\left(\frac{{(-\frac{3}{4})}^n+\left(1\right)^n}{5}\right)^{\frac{1}{n}}=\frac{\left({(-\frac{3}{4})}^n+1\right)^\frac{1}{n}}{5^\frac{1}{n}}{} \to \frac11 = 1$$

Indeed:
if $\forall a_n \to 1$
$$(a_n)^{\frac1n}=e^{\frac{\log a_n}{n}}\to e^0=1$$
and
$$(5)^{\frac1n}=e^{\frac{\log 5}{n}}\to e^0=1$$

and more in general

if $\forall a_n \sim n^k$
$$(a_n)^{\frac1n}=e^{\frac{\log a_n}{n}}\sim e^{k\frac{\log n}{n}}\to e^0=1$$

EG
http://www.wolframalpha.com/input/?i=lim+(n%5E1000000000)%5E(1%2Fn)+n-%3E%2Binf
A: Using the squeeze theorem:
$$
4\leftarrow\left(\frac{4^{n}}{5}\right)^{\frac{1}{n}}=\left|\left(\frac{\left(-4\right)^{n}}{5}\right)^{\frac{1}{n}}\right|\leq\left|\left(\frac{3^{n}+\left(-4\right)^{n}}{5}\right)^{\frac{1}{n}}\right|\leq\left(\frac{2\cdot4^{n}}{5}\right)^{\frac{1}{n}}\rightarrow4
$$
