Find a value for "c" 
For what value of $c$ is $$\lim_{x\to\infty} \left(\frac{x+c}{x-c}\right)^x = e?$$

I am unsure of how to start this question in any sense.
 A: Here's a hint:
$$e = \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x.$$
Let's let $y=x-c$ in your expression (so that $x = y+c$), then since $x\to\infty$, $y\to\infty$ and we get
$$\lim_{x\to\infty}\left(\frac{x+c}{x-c}\right)^x = \lim_{y\to\infty}\left(\frac{y+c+c}{y}\right)^{y+c} = \lim_{y\to\infty}\left(1+\frac{2c}{y}\right)^y\left(1+\frac{2c}{y}\right)^c.$$
Can you take it from here?
A: Hint: If you take logarithm on both sides, you get
$\operatorname{ln}(\lim_{x\to\infty}(\frac{x+c}{x-c})^x)=\operatorname{ln}(e)$.
What do you know about the function $\operatorname{ln}$?
A: You also have a solution using Taylor series.
Let $$A=\left(\frac{x+c}{x-c}\right)^x$$ $$\log(A)=x \log \left(\frac{x+c}{x-c}\right)=x \log \left(\frac{1+\frac cx}{1-\frac cx}\right)$$ For small values of $y$, we also have $$\log \left(\frac{1+y}{1-y}\right)=2 \Big(\frac{y}{1}+\frac{y^3}{3}+\cdots\Big)$$ Replace $y$ by $\frac{c}{x}$ and so $$\log(A)=2x \Big(\frac{ c}{x}+\frac{ c^3}{3 x^3}+O\left(\left(\frac{1}{x}\right)^4\right))=2 c+\frac{2 c^3}{3 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)$$ $$A=e^{2 c}\Big(1+\frac{2 c^3 }{3 x^2}+O\left(\left(\frac{1}{x}\right)^4\right)\Big)$$ which shows the limit and how the expression goes to the limit.
