What is the value of $a$ in this limit? $ \lim_{x\to\infty}\left(\frac{ax-1}{ax+1}\right)^x=9 $ 
What is the value of $a$ in this limit?
  $$ \lim_{x\to\infty}\left(\frac{ax-1}{ax+1}\right)^x=9 $$

I'm preparing for an exam and found this question on a list, but I don't even know how to start. 
Could someone help me?
 A: Hint:
$$ \lim_{x\to\infty}\left(\frac{ax-1}{ax+1}\right)^x=\lim_{x\to\infty}\left( 1+\frac{-2}{ax+1} \right)^{\frac{ax+1}{-2}\cdot\frac{-2x}{ax+1}} $$
Further hint:  

 $$ \lim_{x\to\infty}\left( 1+\frac{-2}{ax+1} \right)^{\frac{ax+1} {-2}\cdot\frac{-2x}{ax+1}}=e^{-2/a} $$

A: This is a standard limit of the form $1^\infty$. 
It is evaluated as $$L=\lim_{x \to \infty}e^{(f(x)-1)(g(x)}$$ where $f(x)$ is the numerator function and $g(x)$ is the denominator function. You can prove this by taking log of the limit. 
So applying it gives the answer as $$L=e^{\frac{-2}{a}}$$. Since $L=9$ so $$a=\frac{-2}{ln9}$$.This question is helpful
A: $\begin{array}\\
\lim_{x\to\infty}\left(\dfrac{ax-1}{ax+1}\right)^x
&=\lim_{x\to\infty}\left(\dfrac{1-1/(ax)}{1+1/(ax)}\right)^x\\
&=\lim_{x\to\infty}\dfrac{(1-1/(ax))^x}{(1+1/(ax))^x}\\
&=\lim_{x\to\infty}\dfrac{((1-1/(ax))^{ax})^{1/a}}{((1+1/(ax))^{ax})^{1/a}}\\
&=\dfrac{e^{-1/a}}{e^{1/a}}\\
&=e^{-2/a}\\
\end{array}
$
So if this equals $9$,
then
$e^{-2/a} = 9$
or
$e^{-1/a} = 3$
so
$-1/a = \ln(3)
$
or
$a = \dfrac{-1}{\ln(3)}
$.
A: Note that $$\lim_{x\to\infty}\left(1+\frac{t}{x}\right)^x=e^t$$  Hence, we have
$$\begin{align}
\lim_{x\to\infty}\left(\frac{ax-1}{ax+1}\right)^x&=\lim_{x\to\infty}\left(\frac{1+\frac{-1/a}{x}}{1+\frac{1/a}{x}}\right)^x\tag1
&=e^{-2/a}
\end{align}$$
Now, set the right-hand side of $(1)$ to $9$ and solve for $a$ to find $a=\frac1{\log(1/3)}$.
