Limit question with exponents, do not understand

Question

I have a function I am trying to simplify. I got it down the final step and I think I am just missing something simple.

I want to show that

$$\underset{n \rightarrow \infty}{\lim}\left( \left (1-\frac{\sqrt{n}t}{n}\right ) ^{-n} \exp(-\sqrt{n}t)\right) = \exp(t^2/2)$$

When I did this problem I used the property: if $a_n$ is a sequence that converges to $a$ as $n$ goes to $\infty$ then

$$\underset{n \rightarrow \infty}{\lim} \left (1+\frac{a_n}{n}\right ) ^{n} = \exp(a)$$

But this means that

$$\underset{n \rightarrow \infty}{\lim} \left (1-\frac{\sqrt{n}t}{n}\right ) ^{-n} = \exp(\sqrt{n}t)$$

Which means that

$$\underset{n \rightarrow \infty}{\lim}\left( \left (1-\frac{\sqrt{n}t}{n}\right ) ^{-n} \exp(-\sqrt{n}t)\right) = 1 \neq \exp(t^2/2)$$

Answer 1

I don't think your theorem can be applied in this case (because the sequence {$-\sqrt(n)t$} doesn't converge). Alternatively,

$$ (1- \dfrac{t}{\sqrt n})^{-n} = e^{ln {(1- \dfrac{t}{\sqrt n})^{-n}}} = e^{-n ln(1-\dfrac{t}{\sqrt n})}$$.

So,$$ (1- \dfrac{t}{\sqrt n})^{-n}. e^{-\sqrt{n}t} = e^{-n ln(1-\dfrac{t}{\sqrt n})-\sqrt{n}t}$$

Considering the taylor series expansion of $ln(1-\dfrac{t}{\sqrt n})$ (about 0), we get,

$$-n(ln(1-\dfrac{t}{\sqrt n})) =-n[ \dfrac{-t}{\sqrt n} - \dfrac{t^2}{ 2n} - \dfrac{t^3}{3 n\sqrt n} ]+ o(t^4 / \sqrt n)$$.

Now, $$-n(ln(1-\dfrac{t}{\sqrt n})) - \sqrt nt = t^2/2+ \dfrac{t^3}{3 \sqrt n} + o(t^4/ \sqrt n).$$

Letting $n \to \infty$, we have $$-n(ln(1-\dfrac{t}{\sqrt n})) - \sqrt nt \to t^2/2 .$$ Hence, $$\lim_{n \to \infty} (1- \dfrac{t}{\sqrt n})^{-n} . e^{- \sqrt nt} = \lim_{n \to \infty} e^{-n(ln(1-\dfrac{t}{\sqrt n})) - \sqrt nt} = e^{t^2/2} $$

Answer 2

The problem in your argument is when you say $$\underset{n \rightarrow \infty}{\lim} \left (1-\frac{\sqrt{n}t}{n}\right ) ^{-n} = \exp(\sqrt{n}t)$$ This doesn't make a lot of sense. $n$ is simply an indexing variable that goes from $1$ to $\infty$. You might as well replace $n$ by $m$ or $k$. Whereas the limit has to equal a specific number for a given $t$.

As for showing the limit, we have $$\begin{align} \left(1-\frac{\sqrt{n}t}{n}\right)^{-n} &= \left(1-\frac{t}{\sqrt{n}}\right)^{-n} \\ &=\exp\left(-n\log\left(1-\frac{t}{\sqrt{n}}\right)\right) \\ &=\exp\left(-n\left(-\frac{t}{\sqrt n}-\frac{t^2}{2n}+O\left(\frac{1}{n^{3/2}}\right)\right)\right) \\ &=\exp\left(t\sqrt{n}+\frac{t^2}{2}+O\left(\frac{1}{n^{1/2}}\right)\right) \\ &=\exp\left(t\sqrt{n}+t^2/2\right)\left(1+O\left(\frac{1}{n^{1/2}}\right)\right) \end{align} $$ Thus, $$ \left(1-\frac{\sqrt{n}t}{n}\right)^{-n}\exp\left(-\sqrt{n}t\right)=\exp\left(t^2/2\right)\left(1+O\left(\frac{1}{n^{1/2}}\right)\right) $$ and the limit is what you claim.

P.S.: Your limit is an indeterminate form $0\times\infty$. This means that you cannot split it into the product of $2$ limits. You have to consider the entire expression.

Answer 3

For simplification, let $n=m^2$ to make $$\underset{n \rightarrow \infty}{\lim}\left( \left (1-\frac{\sqrt{n}t}{n}\right ) ^{-n} e^{-\sqrt{n}t}\right) =\underset{m \rightarrow \infty}{\lim}\left( \left(1-\frac{t}{m}\right)^{-m^2}e^{-m t}\right)$$ Now $$A=\left(1-\frac{t}{m}\right)^{-m^2}\implies \log(A)=-m^2\log\left(1-\frac{t}{m}\right)$$ Using Taylor expansion, we then have $$\log(A)=-m^2\left(-\frac{t}{m}-\frac{t^2}{2 m^2}-\frac{t^3}{3 m^3}+O\left(\frac{1}{m^4}\right)\right)=m t+\frac{t^2}{2}+\frac{t^3}{3 m}+O\left(\frac{1}{m^2}\right)$$ $$A=e^{\log(A)}=\exp \left(mt +\frac{t^2}{2}+\frac{t^3}{3 m}+O\left(\frac{1}{m^2}\right)\right)$$ $$A \,e^{-mt}=e^{\frac{t^2}{2}}\left(1+\frac{t^3}{3 m}+O\left(\frac{1}{m^2}\right) \right)=e^{\frac{t^2}{2}}\left(1+\frac{t^3}{3 \sqrt n}+O\left(\frac{1}{n}\right) \right)$$ which shows the limit and how it is approached.