calculate the limit of a function I want to calculate the limite of this function when $x\to\infty$.
$\lim_{x\to\infty}\left(\frac{c+\sqrt{x}}{-c+\sqrt{x}}\right)^x\exp(-2c\sqrt{x})$, where $c$ is a constant.
Numerically, I plot a graphic of this function, and I think the answer is 1. But theoretically, I have no idea how to proceed.
 A: Using $\log\left(\frac{1+x}{1-x}\right)=\sum_{n=0}^\infty \frac{2x^{2n+1}}{2n+1}$ we find
$$\begin{align}
\left(\frac{c+\sqrt{x}}{-c+\sqrt{x}}\right)^x e^{-2c\sqrt x}&=e^{x\log\left(\frac{1+c/\sqrt{x}}{1-c/\sqrt{x}}\right)}e^{-2c\sqrt x}\\\\
&=\exp\left({2cx^{1/2}\sum_{n=1}^\infty \frac{c^{2n}}{(2n+1)x^{n}}}\right)
\end{align}$$
Letting $x\to \infty$, we find
$$\lim_{x\to\infty}\left(\frac{c+\sqrt{x}}{-c+\sqrt{x}}\right)^x e^{-2c\sqrt x}=1 $$
A: If you want to go beyond the limit, considering
$$y= \left(\frac{\sqrt{x}+c}{\sqrt{x}-c}\right)^x \,e^{-2 c \sqrt{x}}$$ take logarithms
$$\log(y)=x \log\left(\frac{\sqrt{x}+c}{\sqrt{x}-c}\right)-2 c \sqrt{x}=x \log \left(1+\frac{2 c}{\sqrt{x}-c}\right)-2 c \sqrt{x}$$ Now use the series expansion of $\log(1+\epsilon)$ with $\epsilon=\frac{2 c}{\sqrt{x}-c}$ and continue with long division to obtain
$$\log(y)=\frac{2 c^3}{3 \sqrt{x}}\Bigg[1+\frac{3 c^2}{5 x}+O\left(\frac{1}{x^2}\right) \Bigg]$$ that is to say
$$y \sim \exp\Bigg[\frac{2 c^3}{3 \sqrt{x}} \Bigg]$$
