Evaluate $\lim_{n \to \infty}e^{-n\sinh^{-1}(T/n)} $ According to Wolfram Alpha $\lim_{n \to \infty}e^{-n\sinh^{-1}(T/n)} = e^{-T}.$  I would like to prove this.
Here is what I have tried so far
$e^{-n\sinh^{-1}(T/n)} = e^{-n\log({(T/n) + \sqrt{(1 + T^2/n^2)}})} = (T/n + \sqrt{(1 + T^2/n^2)})^{-n}.$
From here it suffices to show that $T/n + \sqrt{(1 + T^2/n^2} = 1 + T/n$ since then
$\lim_{n \to \infty}e^{-n\sinh^{-1}(T/n)} = \lim_{n\to \infty}(1 + T/n)^{-n} = \lim_{n \to \infty}\frac{1}{(1 + T/n)^n} = e^{-T}.$
Any help is appreciated.
 A: You don't have $T/n+\sqrt{1+T^2/n^2}=1+T/n$ but:
$$\frac{T}{n}+\sqrt{1+\frac{T^2}{n^2}}=1+\frac{T}{n}+O\left( \frac{1}{n^2} \right)$$
in fact you can directly use the first order expansion of $\sinh^{-1}$:
$$\sinh^{-1}\left( \frac{T}{n} \right)=\frac{T}{n}+O\left( \frac{1}{n^2} \right)$$
so:
$$e^{n\sinh^{-1}\left( \frac{T}{n} \right)}=\exp\left(-T+O\left( \frac{1}{n} \right)\right)=e^{-T} e^{o(1)} \longrightarrow_{n \to \infty} e^{-T}$$
A: It would suffice, but it's not true: you cannot prove it, as it'd be equivalent to showing
$$
\sqrt{1+\frac{T^2}{n^2}} = 1
$$
for all $T,n$ (which is clearly false). 
Let's try something different, then. Fixing $T$, we want to compute the limit of
$$
e^{-n\log\left( \frac{T}{n} + \sqrt{1+\frac{T^2}{n^2}}\right)}
$$
and by continuity of $\exp$ what we want is equivalent to showing that
$$
\lim_{n\to\infty} n\log\left( \frac{T}{n} + \sqrt{1+\frac{T^2}{n^2}}\right) = T\,.\tag{1}
$$
Since $\lim_{n\to\infty} \frac{T}{n} = 0$, we can use the Taylor expansion of $\sqrt{1+u}$ for $u$  around $0$ to write
$$
n\log\left( \frac{T}{n} + \sqrt{1+\frac{T^2}{n^2}}\right)
= n\log\left( \frac{T}{n} + 1+\frac{T^2}{2n^2} + o\left(\frac{1}{n^2}\right)\right)
$$
and using the Taylor series of $\log(1+u)$ around $0$, to get from there
$$
n\log\left( 1+ \frac{T}{n} +\frac{T^2}{2n^2} + o\left(\frac{1}{n^2}\right)\right)
= n\left(\frac{T}{n} + \frac{T^2}{2n^2} - \frac{T^2}{2n^2}+ o\left(\frac{1}{n^2}\right) \right)
= T + o\left(\frac{1}{n}\right) \xrightarrow[n\to\infty]{} T \tag{2}
$$
establishing  (1).
A: Take $m=\frac{1}{n}$, then you have
$$e^{\lim_{m\to 0+} -\frac{\log (Tm+\sqrt{1+T^2m^2})}{m}}$$
Now we use good old L'Hopital inside the exponent to get
$$\lim_{m\to 0+} -\frac{\log (Tm+\sqrt{1+T^2m^2})}{m}=\lim_{m\to 0+}-\frac{T+(1+T^2m^2)^{-1/2}T^2m}{Tm+\sqrt{1+T^2m^2}}$$
Fortunately, lots of things go to zero, leaving the limit as just $-T$.
