Finding $\lim_{x\to\infty} \frac{(2x-5)^4}{(2x^2+1)(3x^2-2)}$ Finding $$\lim_{x\to\infty} \frac{(2x-5)^4}{(2x^2+1)(3x^2-2)}$$
Do I multiply top & bottom by $\frac{1}{x^2}$ or $\frac{1}{x^4}$? How do I distribute them into the numerator tho? 
In the answer given, I think with some typos: 
$$... = \lim_{x\to\infty} \frac{(2-5/x)^\color{red}2}{(2+1/x^2)(3-2/x^2)} = \frac{2^{\color{red}4}}{2\cdot 3} = \frac{8}{3}$$
Either way, how do I distribute the $\frac{1}{x^?}$ into the numerator? 

The dominant term is $x^4$ (Imagine that you expanded both the numerator and denominator. What would the highest power of $x$ be?). So, you could either multiply numerator and denominator by $1/x^4$ (see below), or factor as follows:
$$\eqalign{
{(2x-5)^4 \over (2x^2+1)(3x^2-2)}
&={ \bigl(x(2-{5\over x})\bigr)^4\over     x^2 (2+{1\over x^2}) \cdot x^2(3-{2\over x^2 } )   }\cr
&={x^4 (2-{5\over x})^4 \over  x^2(2+{1\over  x^2})\cdot x^2(3-{2\over  x^2  })}\cr
&={(2-{5\over x})^4\over  (2+{1\over x^2})(3-{2\over  x^2})}.
}
$$
The limit as $x$ tends to infinity is $$ {2^4\over 2\cdot 3}=16/6 =8/3.$$
The $(2-5/x)^2$ is probably a typo..

Using the other approach, where you multiply numerator and denominator by $1/x^4$:
In the numerator, to distribute $1/x^4$ over $(2x-5)^4$: 
$$\textstyle
{1\over x^4}(2x-5)^4 = \bigl(  {1\over x} (2x-5) \bigr)^4=(2-{5\over x})^4.
$$
(just using $a^nb^n=(ab)^n$ here).
In the denominator, you could do the following:
$$\textstyle
{1\over x^4}(2x^2+1)(3x^2-2)
={1\over x^2}(2x^2+1) {1\over x^2}(3x^2-2)= (2+{1\over x^2})(3-{2\over x^2})
$$
(or just expand $(2x^2+1)(3x^2-2)$  first and then distribute the $1/x^4$ across).
