Limit of $(1+\frac{1}{n^{3\alpha}})^{n^{5}}$ I was trying to solve the $\lim\limits_{n \to \infty}(1+\frac{1}{n^{3\alpha}})^{n^{5}}$
where I have to say for which $\alpha$ parameter the limit  is finite.
I tried to sobstitute $t=n^{3\alpha}$:
$(1+\frac{1}{t})^{{t}^{\frac{5}{3\alpha}}}$
and I found $\alpha \ge 0$
But the solution says  $\alpha\geq \frac{5}{3}$
Where I'm wrong?
 A: First exclude the case $\alpha\le0$. For $\alpha>0$, compute instead
$$
\lim_{x\to\infty}\left(1+\frac{1}{x^{3\alpha}}\right)^{x^5}
$$
(the function as opposed to the sequence). As a preparation, we compute the limit of the logarithm, but substituting $x=1/t$:
$$
\lim_{t\to0^+}\frac{\log(1+t^{3\alpha})}{t^5}=
\lim_{t\to0^+}\frac{t^{3\alpha}+o(t^{3\alpha})}{t^5}
$$
and you want the limit is finite. This clearly means $3\alpha\ge 5$.
Fill in the details.
A: $$\left(1+\frac1{n^{3\alpha}}\right)^{n^5}=\left[\left(1+\frac1{n^{3\alpha}}\right)^{n^{3\alpha}}\right]^{n^{5-3\alpha}}$$
Now, for $\;3\alpha>0\;$ we get $\;n^{3\alpha}\xrightarrow[n\to\infty]{}\infty\;$ , so
$$\left(1+\frac1{n^{3\alpha}}\right)^{n^{3\alpha}}\xrightarrow[n\to\infty]{}e$$, and if $\;5+3\alpha>0\;$ then
$$\left[\left(1+\frac1{n^{3\alpha}}\right)^{n^{3\alpha}}\right]^{n^{5-3\alpha}}\ge 2^{n^{5-3\alpha}}\xrightarrow[n\to\infty]{}\infty$$
so we must have
$$5-3\alpha\le0\iff\alpha\ge\frac53$$
A: Starting from what you did:
First of all, for $\alpha\leq 0$ you have $1+\frac{1}{n^{3\alpha}}\geq 2$, and $2^{n^5}$ diverges to $\infty$. This settles this case, so now we consider $\alpha > 0$.
Consider what you have:
$$\begin{align}
\left(1+\frac{1}{t}\right)^{{t}^{\frac{5}{3\alpha}}}
 &= \exp\left({t}^{\frac{5}{3\alpha}} \ln\left(1+\frac{1}{t}\right)\right)
 = \exp\left(\frac{{t}^{\frac{5}{3\alpha}}}{t}\cdot t \ln\left(1+\frac{1}{t}\right)\right)
\\&= \exp\left({t}^{\frac{5-3\alpha}{3\alpha}}\cdot t \ln\left(1+\frac{1}{t}\right)\right)
\end{align}$$
Recalling that $t \ln\left(1+\frac{1}{t}\right)\xrightarrow[t\to\infty]{}1$, we get:


*

*If $5-3\alpha < 0$, then the exponent goes to $0\cdot 1=0$

*If $5-3\alpha = 0$, then the exponent goes to $1\cdot 1=1$

*If $5-3\alpha > 0$, then the exponent goes to $\infty\cdot 1=\infty$


Can you conclude?
