Find the limit... Find $\lim _{x\to \infty \:}\left(\frac{x^2+2}{x^2+1}\right)^{3x^2+\frac{1}{x}}$
using that $\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)^x=e.$
I'm trying to divide up the limit into a bunch of separate pieces and then substitute n=to something but I'm stuck on how to manipulate the original to be able to use the given limit. 
 A: $$\lim _{x\to \infty \:}\left(\frac{x^2+2}{x^2+1}\right)^{3x^2+\frac{1}{x}} = \lim _{x\to \infty \:} \left(1 + \frac 1{x^2+1}\right)^{3(x^2+1)} \times \lim _{x\to \infty \:} \left(1 + \frac 1{x^2+1}\right)^{\frac 1x - 3}$$
Now since $x^2 + 1 \to \infty$ as $x \to \infty$ you'll get $e^3$ for the first factor. While for the second factor substitute $x=\infty$ and you'll get $1^{-3}=1$
A: $$\lim_{x\to\infty}\left(\frac{x^2+2}{x^2+1}\right)^{3x^2+\frac{1}{x}}=\lim_{x\to\infty}\left({1+\frac{1}{x^2+1}}\right)^{(x^2+1)\cdot\frac{{3x^2+\frac{1}{x}}}{x^2+1}}=e^{\lim_{x\to\infty}\frac{3x^3+1}{x^3+x}}=e^3$$
A: this can be done too using the fact $$(1+small)^{BIG} = e^{small \times BIG}$$ which  follows from the definition $\lim_{n \to \infty} (1 + 1/n)^n = e$ of $e$ and the additive properties of the exponential function.
$$(x^2 + 2)(x^2 + 1)^{-1} = (x^2 + 2)(x^{-2} -x^{-4}+\cdots) = 1+x^{-2}+\cdots $$
therefore $$\left(\frac{x^2 + 2}{x^2 + 1}\right)^{3x^2 + \frac1x}=e^{x^{-2}(3x^2 + \frac1x + \cdots)} = e^3 + \cdots$$  and 
$$\lim_{x \to \infty}\left(\frac{x^2 + 2}{x^2 + 1}\right)^{3x^2 + \frac1x} = e^3.$$
