What is $\lim_{n\to\infty}\left(\frac{2}{5}\right)^{1/n}$? What is $$\lim\limits_{n\to\infty}\left(\frac{2}{5n}\right)^{\frac{1}{n}}$$ This is clearly equal to $\lim\limits_{n\to\infty}\left(\frac{2}{5}\right)^{\frac{1}{n}}\left(\frac{1}{n}\right)^{\frac{1}{n}}$. We know that $\lim\limits_{n\to\infty}\left(\frac{1}{n}\right)^{\frac{1}{n}}=1$. What is $\lim\limits_{n\to\infty}\left(\frac{2}{5}\right)^{\frac{1}{n}}$, though?
 A: Hint:


*

*Convert the power to logarithmic form ($a^b = e^{b*\log{a}}$).

*Use the L'Hopitals rule.
A: We have $$\left(\frac{2}{5n}\right)^{\frac{1}{n}} = e^{\frac{1}{n} \log \left(\frac{2}{5n}\right)}$$ and $$\frac{1}{n} \log \left(\frac{2}{5n}\right) = \frac{\log2 - \log5 - \log n}{n} \to 0$$ Thus $$\lim\limits_{n\to\infty}\left(\frac{2}{5n}\right)^{\frac{1}{n}} = 1$$
A: The one you ask for is easier than the one you know:
$$
\bigl(\tfrac{2}{5}\bigr)^{1/n}=e^{\tfrac{1}{n}\ln(2/5)}.
$$
Now, $\frac{1}{n}\ln(2/5)\to 0$ as $n\to+\infty$, and the exponential function is continuous at zero. Thus, the limit is $e^0=1$.
A: As $n$ gets large, $\frac{1}{n}$ gets smaller, nearing $0$. Then $(\frac{2}{5})^{\frac{1}{n}}$ has an exponent that tends to $0$, and you probably know what something to the $0$th power is.
A: $$\lim_{n\rightarrow\infty}\Big(\frac{2}{5n}\Big)^{\frac{1}{n}}=\lim_{\frac{5n}{2 }\rightarrow\infty}\Big(\frac{2}{5n}\Big)^{\frac{2}{5n}\frac{5}{2}}=1^{\frac{5}{2}}=1 \mbox{ (since region of interest is }\Bbb R_{\geq0}\mbox{)}.$$
