Find $\lim_{n\rightarrow\infty}\left(1-(1-\exp(tn^{-\frac{1}{v}}))^v\right)^n$ Find the limit as $n\rightarrow\infty$ of
$\left(1-(1-\exp(tn^{-\frac{1}{v}}))^v\right)^n$,
where $t\in(-\infty,0)$, and $v\in(0,1)$.
Remarks:
A non-trivial limit does exist! - verified numerically.
I would like to use a similar idea to $\lim_{n\rightarrow\infty}\left(1-\frac{t}{n}\right)^n=\exp(-t)$. This standard result can be proved, for example, by taking the logarithm and using l'Hopitals rule. The method does not seem to work in this case however due to problems differentiating.
 A: Putting $r=-t>0$ and $u=v^{-1}>1$ your expression equals to:
\begin{align*}
\left(1-\left(1-\frac 1{e^{\frac r{n^{u}}}}\right)^v\right)^n
=&\left(1-\frac1{\left(1-\frac 1{e^{\frac r{n^{u}}}}\right)^u}\right)^n\;\;.
\end{align*}
Now it's clear that
$$
\frac r{n^{u}}\stackrel{n\to+\infty}{\longrightarrow}0^+
$$
thus
$$
1-\frac1{\left(1-\frac 1{e^{\frac r{n^{u}}}}\right)^u}\stackrel{n\to+\infty}{\longrightarrow}-\infty$$
thus
$$
\lim_{n\to+\infty}\left(1-\frac1{\left(1-\frac 1{e^{\frac r{n^{u}}}}\right)^u}\right)^n=\infty\;.
$$
A: Instead of a sequence, consider the limit of the function
$$
\lim_{x\to\infty}\left(1-(1-\exp(tx^{-\frac{1}{v}}))^v\right)^{x}
$$
Set $y=1/x$ and try computing the limit of the logarithm, that is
$$
\lim_{y\to0^+}\frac{\log(1-(1-\exp(ty^{1/v}))^v)}{y}
$$
If it exists, it is the derivative at $0$ of the function
$$
f(y)=\log(1-(1-\exp(ty^{1/v}))^v)
$$
Now the derivative can be explicitly computed:
$$
f'(y)=
\frac{1}{1-(1-\exp(ty^{1/v}))^v}\cdot
v(1-\exp(ty^{1/v}))^{v-1}\cdot
(-\exp(ty^{1/v}))\cdot\frac{t}{v}y^{(1-v)/v}
$$
However, this does not exist at $0$, because of the term
$$
(1-\exp(ty^{1/v}))^{v-1}
$$
which is not defined at $0$. However l'Hôpital allows us to go on.
The first factor has limit $1$ as well as $\exp(ty^{1/v})$. The factors $v$ cancel out, so we are left with $-t$ times
$$
\lim_{y\to0^+}(1-\exp(ty^{1/v}))^{v-1}y^{(1-v)/v}=
\lim_{y\to0^+}\frac{y^{(1-v)/v}}{(1-\exp(ty^{1/v}))^{1-v}}
$$
By setting $z=y^{1/v}$, we have
$$
\lim_{z\to0^+}\frac{z^{1-v}}{(1-\exp(tz))^{1-v}}=
\lim_{z\to0^+}\left(\frac{1-\exp(tz)}{z}\right)^{v-1}=(-t)^{v-1}
$$
So finally the limit is $(-t)^v$ and your big initial limit is
$$
\exp((-t)^v)
$$
