Limit of Functions Using $\lim_{n\to\infty} (1+\frac{x}{n}) ^n = e^x $ Can someone help me solve the following three exercices? 
1) $\displaystyle \lim_{x\to \infty} \left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{x^2 + \Large\frac{8}{x} } $
2) $\displaystyle \lim_ {x \to \infty} \left( 1+ \frac{1}{9x^4 + 8x^2 +8 }\right) ^{8x^4 +9x^2 + 1}  $
3) $\displaystyle \lim_{x\to 0 } \sqrt[\Large x]{1+9x}$ 
These aren't homework, but pre-exam (not formal) exercices
Thanks in advance

I'll show you one, but they are all very similar:
$$\begin{align*}\lim_{x\to \infty} \left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{x^2 + \large \frac{8}{x} }&=\lim_{x\to \infty} \left(\left( 1+ \frac{1}{9x^2 + x + \frac{1}{x} } \right)^{9x^2+x+\Large \frac1x} \right)^{\Large \frac{x^2 + \Large\frac{8}{x}}{9x^2+x+\frac1x} }\overset{(*)}{=}\\
&=\lim_{x\to \infty}e^{\Large \frac{x^2 + \frac{8}{x}}{9x^2+x+\frac1x} }=\lim_{x\to \infty}e^{\Large \frac{1 + \frac{8}{x^3}}{9+\frac1x+\frac1{x^3}} }=e^{1/9}\end{align*}$$
Where $(*)$ follows from the fact that $\lim_{x\to\infty}9x^2+x+\frac1x=\infty$.
Can you follow the solution? Can you apply it to the other two?
