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
 A: There is a simple way to solve $1^\infty$ type limits. Suppose you need to find $\lim f(x)^{g(x)}, f(x) \to 1, g(x)\to \infty$
Let $$P = f(x)^{g(x)}$$
Define $$p = \log{P} = g(x)\log{f(x)}$$
As $f$ is close to one, its logarithm can be expanded using the standard Taylor series for logs.
$$\log(1+x) = x -\frac{x^2}{2}+\ldots$$
So, $$\log(f(x))\approx f(x)-1 - \frac{(f(x)-1)^2}{2}+\ldots$$
Usually the first term is enough.
So,
$$p\approx g(x)(f(x)-1) + \mbox{higher order terms}$$
So,
$$P = \lim_{x\to x_0} e^{g(x)(f(x)-1)}$$
Applying it to your case,
1. $$p = g(x)(f(x)-1) = (x^2 + \frac{8}{x})(\frac{1}{9x^2+\ldots})$$
$$\lim_{x \to \infty} p = \frac{1}{9}$$
So, your final answer is $P = e^p = e^{\frac{1}{9}}$
For 2, you can tell the answer by inspection. $P = e^{\frac{8}{9}}$
For 3, you need $$p = \lim_{x \to 0}\frac{\sqrt{1+9x}-1}{x}$$
Binomial expansion or L'Hospital Rule gives $p = \frac{9}{2}$
So, your answer is $P = e^{\frac{9}{2}}$
A: 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?
A: For example:
$$\sqrt[x]{1+9x}=(1+9x)^{1/x}=\left(1+\frac{9}{1/x}\right)^{1/x}\xrightarrow[x\to 0^+]\ldots$$
Just take into consideration that the limit in this case must be $\,x\to 0^+\,$ : from the right.
