Limit of over the real line I am trying to find $$ \lim_{x \to \infty} \biggl(\frac{x + \log9}{x - \log9} \biggr)^x $$
While I know the methods for getting limits of various indeterminate forms like like $\frac{0}{0}, \frac{\infty}{\infty},1^\infty,0\times0,\infty \times \infty$ this appears to be $\bigl(\frac{\infty}{\infty}\bigr)^\infty$.
How to convert it to some other indeterminate form whose limit can be found by standard methods. 
 A: By a shift of the variable
$$ \lim_{x \to \infty} \biggl(\frac{x + \log9}{x - \log9} \biggr)^x =\lim_{y \to \infty} \biggl(\frac{y +2 \log9}{y} \biggr)^{y+\log 9} =\lim_{y \to \infty} \biggl(1+\frac{2 \log9}{y} \biggr)^y\biggl(1+\frac{2 \log9}{y} \biggr)^{\log 9}.$$
The first factor is recognized as a case of the exponential function,
$$\lim_{y \to \infty} \biggl(1+\frac{2 \log9}{y} \biggr)^y=\lim_{z \to \infty} \biggl(1+\frac1z \biggr)^{z\,2 \log9}=e^{2\log 9}=\color{green}{81}$$ while the second factor tends to $1$.
A: Let $$f(x) = \left( \frac{x+\log 9}{x-\log 9} \right)^x.$$ We want to compute $\lim_{x\to\infty} \log f(x)$, and then exponentiate the result.
$$
\lim_{x\to\infty} x \log \frac{x+\log 9}{x-\log 9} = \lim_{x\to\infty} \frac{\log \frac{x+\log 9}{x-\log 9}}{\frac{1}{x}} = \lim_{x\to\infty}\frac{(2\log 9) x^2}{x^2 - \log^2 9} = 2\log 9,
$$
where the second equality follows from L'Hospital Rule. The final answer should be $e^{2\log 9} = 81$.
A: this limit in the ${ 1 }^{ \infty  }$ form so
$$\lim _{ x\to \infty  } \left( \frac { x+\log  9 }{ x-\log  9 }  \right) ^{ x }=\lim _{ x\to \infty  } \left( 1+\frac { 2\log  9 }{ x-\log  9 }  \right) ^{ x }=\lim _{ x\to \infty  }{ \left[ \left( 1+\frac { 2\log  9 }{ x-\log  9 }  \right) ^{ \frac { x-\log  9 }{ 2\log  9 }  } \right]  } ^{ \frac { 2\log  9 }{ x-\log  9 } x }=\\ =\lim _{ x\to \infty  }{ e } ^{ \frac { 2\log  9 }{ x-\log  9 } x }={ e }^{ 2\log  9 }=81$$
A: Write  $$\biggl(\frac{x + \log9}{x - \log9} \biggr)^x =e^{x\ln{(\frac{x + \log9}{x - \log9})}}$$
I have not checked if the limit can be solved with this form but this is a form you know and you can try working with it.
If you want you can solve the limit $$\lim_{x \rightarrow +\infty}\ln{(\frac{x + \log9}{x - \log9})}$$
$P.S:$
This answer was downvoted before some minutes but i do not understand why,because the O.P did not ask  for his limit to be solved completely.
