Evaluate $\lim\limits_{n\to\infty}(\frac{3}{x^{4n}}+1)^\frac{1}{n},|x|<1,n\in\mathbb{N}$ The limit is equal to $1$. I don't understand how.
If $n\rightarrow \infty$ and $|x|<1$, for example if $x=0.5$ we have
$$\frac{3}{0.5^\infty}$$ which I think should be $\infty$?
 A: You can squeeze as follows:
$$\bigg(\frac{3}{x^{4n}}\bigg)^\frac{1}{n} < \bigg(\frac{3}{x^{4n}}+1\bigg)^\frac{1}{n} < \bigg(\frac{4}{x^{4n}}\bigg)^\frac{1}{n}$$
This is the same as
$$\frac{3^{1 \over n}}{x^{4}} < \bigg(\frac{3}{x^{4n}}+1\bigg)^\frac{1}{n} < \frac{4^{1 \over n}}{x^{4}}$$
Both the left and the right sides of the above go to ${\displaystyle {1 \over x^4}}$ as $n$ goes to infinity, so that has to be the limit.
A: If $|x|<1$ we have a limit with indeterminate form $\infty^0$. We can then use L'Hospital's by rewriting the limit as $$\lim_{n \to \infty}e^\dfrac{\ln(1+3x^{-4n})}{n}$$ Use L'Hospital's rule on the exponent. I believe after one application of LR we have the limit $$\lim_{n \to \infty} \frac{-12x^{-4n}\ln(x)}{1+3x^{-4n}} = \lim_{n \to \infty} \frac{-12\ln(x)}{x^{4n}+3}$$ Since $|x|<1$ we know $\lim_{n \to \infty} x^{4n} = 0$, which means
$$ \lim_{n \to \infty} \frac{-12\ln(x)}{x^{4n}+3} \approx \lim_{n \to \infty} \frac{-12\ln(x)}{0+3} = -4\ln(x)$$ 
 Hence  $$\begin{align}\lim\limits_{n\to\infty}\left(\frac{3}{x^{4n}}+1\right)^\frac{1}{n} = e^{-4\ln(x)} \\ = x^{-4} \end{align}$$ With thanks to kobe for catching my mistake. It may also be worth noting that, had $|x|>1$ been the case, then  $\lim_{n \to \infty} \left|x^{4n}\right| = \infty$ so we would have found $$ \lim_{n \to \infty} \frac{-12\ln(x)}{x^{4n}+3} = 0$$ and hence your limit would have been $e^0 = 1$ which would explain your initial confusion.
A: You may write, for $0<|x|<1$, as $n \to +\infty$,
$$
\begin{align}
\left(\frac{3}{x^{4n}}+1\right)^{1/n}&=e^{\large \frac1n \log \left(\frac{3}{x^{4n}}+1\right)}\\\\
&=e^{\large \frac1n \log \left(\frac{3}{x^{4n}}\right)+\large \frac1n \log \left(1+\frac{x^{4n}}3\right)}\\\\
&=e^{\large \frac1n \log 3 -\frac1n\log (x^{4n})+\frac1{3n} x^{4n}+O\left(\frac1{n^2}\right)}\\\\
&=e^{\large -4\log x+\frac1n \left(\log 3+\frac1{3} x^{4n}\right)+O\left(\frac1{n^2}\right)}\\\\
&=\frac1{x^4}e^{\large \frac1n \left(\log 3+\frac1{3} x^{4n}\right)+O\left(\frac1{n^2}\right)}\\\\
& \to \frac1{x^4}.
\end{align}
$$
A: $\dfrac{\ln(3y^n+1)}{n}=\dfrac{\ln 3+n\ln y + \ln(1+\frac{1}{3y^n})}{n}\to \ln y, n \to \infty\Rightarrow f_n(x) \to y = \dfrac{1}{x^4}$
