What's the limit of $(1+\frac{1}{8^n})^n$ What's the limit of $(1+\frac{1}{8^n})^n$? How do I find the answer for this?
Thanks in advance. 
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$$
\pars{1 + {1 \over 8^{n}}}^{n}
=
\expo{n\ln\pars{1 + 8^{-n}}}
\sim
\expo{n/8^{n}} \to \color{#0000ff}{\large 1}
$$
A: Hint: Use the Binomial Theorem:
$$(1+8^{-n})^n=\sum_{k=0}^n 8^{-nk}\binom nk = 1 + 8^{-n}\left(\sum_{k=1}^n 8^{-n(k-1)}\binom nk\right)$$
and the identity $\displaystyle\binom nk = \binom{n-1}{k-1}+\binom{n-1}k$ to estimate the parenthesized sum.

NB. Logarithms (as Steven Stadnicki suggests) provide an easier solution, but perhaps you don't have them available yet.
A: It might be useful to let $L=\lim_{n\to\infty} \left(1+\frac{1}{8^n}\right)^n$ and then notice that $\ln(L)=\lim_{n\to\infty}n\cdot\ln\left(1+\frac{1}{8^n}\right)$. Because $\ln(\cdot)$ is a continuous function, we can do that.  Then, you might be able to solve it from there.
A: $$\lim_{x \to \infty}  (1+\frac{1}{8^n})^n$$
$$\lim_{x \to \infty}((1+\frac{1}{8^n})^{8^n})^\frac{n}{8^n}$$
$$e^{lim_{x \to \infty}\frac{n}{8^n}} = e^0 = 1$$
