Evaluation of $\lim_{m\to\infty}\Big(F(e^{-\frac{\lambda}{m^2}})\Big)^m$ given $F(z)=\frac{1-\sqrt{1-z^2}}{z}$ I am given $F(z):=\frac{1-\sqrt{1-z^2}}{z}$ defined for $z\in\mathbb{R}$ and $|z|<1$, and I would like to show that for every real number $\lambda>0$
\begin{equation}
\lim_{m\to\infty}\Big(F(e^{-\frac{\lambda}{m^2}})\Big)^m=e^{-\sqrt{2\lambda}}
\end{equation}
Below is my unsuccessful attempt
\begin{split}
&{\hspace{6mm}} \lim_{z\to1-}\Big(1-F(z)\Big) = \frac{\sqrt{1-z}(\sqrt{1+z}-\sqrt{1-z})}{z}\approx \sqrt2\sqrt{1-z} \\
& \Rightarrow \lim_{m\to\infty}\Big(F(e^{-\frac{\lambda}{m^2}})\Big)^m =\lim_{m\to\infty} \Bigg (1-\sqrt{2}\sqrt{1-e^{-\frac{\lambda}{m^2}}}\Bigg)^m
\end{split} 
I tried using the Binomial Theorem to expand $\sqrt{.}$ and $(.)^m$, and expanding $e^{-\frac{\lambda}{m^2}}$ to a polynomial. None of these methods work so far.
Related Facts
Let $\tau(m)$ be the first passage time that a simple random walk on the integers reaches the state $m$. The probability generating function for $\tau(1)$ is $Ez^{\tau(1)}=F(z)$. The probability generating function for $\tau(m)$ is $Ez^{\tau(m)}=\Big(F(z)\Big)^m$. The rescaled random variable $\tau(m)/m^2$ converge in distribution to the one-sided stable law of exponent $1/2$.     
 A: Let $z_m=\exp(-\lambda/m^2)$. You have 
$$F(z_m)-1=-\frac{2\sqrt{1-z_m}}{\sqrt{1-z_m}+\sqrt{1+z_m}}$$ and $1-z_m\sim \frac{\lambda}{m^2}$. Hence $F(z_m)-1\sim -\frac{\sqrt{2\lambda}}{m}$. 
Now we take the logarithm, and as $\log(1+x)\sim x$ if $x\to 0$ we get:
$$m\log(F(z_m))=m\log (1+(F(z_m)-1))\sim m(F(z_m)-1)\sim m(-\frac{\sqrt{2\lambda}}{m})=-\sqrt{2\lambda}$$
and we are done. 
A: set $\large{e^{\frac{\lambda}{m^2}}}=t\,$. We have $m=\sqrt{\frac{\lambda}{\ln t}}$. As $m\to +\infty$ then $t\to 1^{+}$
$$F\left(e^{-\frac{\lambda}{m^2}}\right)=F\left(\frac{1}{t}\right)=t-\sqrt{t^2-1}$$
$$I=\lim_{m\to\infty}\Big(F(e^{-\frac{\lambda}{m^2}})\Big)^m=\lim_{t\to1}(t-\sqrt{t^2-1})^{\large {\sqrt{\frac{\lambda}{\ln t}}}}=\exp\left(\lim_{t\to1}\frac{t-1-\sqrt{t^2-1}}{\sqrt{\frac{\ln t}{\lambda}}}\right)$$
$$I=\exp\left(\sqrt{\lambda}\lim_{t\to1^+}\frac{t-1-\sqrt{t^2-1}}{\sqrt{t-1}}\right)$$
$$I=\exp\left(\sqrt{\lambda}\lim_{t\to1^+}\frac{1-\frac{t}{\sqrt{t^2-1}}}{\frac{1}{2\sqrt{t-1}}}\right)=\exp\left(2\sqrt{\lambda}\lim_{t\to1^+}{\sqrt{t-1}-\frac{t}{\sqrt{t+1}}}\right)=\exp\left(2\sqrt{\lambda}\times \frac{-1}{\sqrt{2}}\right)$$
$$I=e^{-\sqrt{2\lambda}}$$
