# Prove: $\lim_{x\rightarrow \infty} \left (\frac{x+\log 9}{x-\log 9} \right )^{x}=81$

Over the real line, prove that

$$\lim_{x\rightarrow \infty} \left (\frac{x+\log 9}{x-\log 9} \right )^{x}=81$$

I've tried L Hospital Rule, which gets too complicated and I'm trying by Squeeze theorem now.

• Do you know that $(1+(a/x))^{x}\to e^a$ as $x\to\infty$. Try to guess $a$ in your question. – Paramanand Singh Nov 7 '18 at 16:37
• @ParamanandSingh $\lim_{(x-\log 9)\rightarrow \infty}\left ( 1+ \frac{2\log9}{x-\log 9} \right )^{\log 9}\left ( 1+ \frac{2\log9}{x-\log 9} \right )^{x-\log 9}=1.e^{2\log9}=81$ – Akash Gaur Nov 8 '18 at 11:49
• Yes that's exactly what you need to do. – Paramanand Singh Nov 8 '18 at 12:15

Hint: $$\frac{x+a}{x-a}=1+\frac{2a}{x-a}$$
Let $$u=x-\log 9$$. As $$x \to \infty$$ we have $$u \to \infty$$.