How do you figure out the:
lim X-> OO (x-3)^8 (2-X)^3/ (X-5)^5 (4-X)
You could divide all of the terms in the expression by the highest term in the denominator. The highest term in the denominator would be $\displaystyle x^6$, and the highest term in the numerator would be $\displaystyle x^{11}$. This way you can find out that there will be a $\displaystyle x^5$ expression left meaning the limit as x approaches infinity is infinity.
Under standard order of operations, this would be $\displaystyle \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3(4-x)}{(x-5)^5},$ but I am guessing that you intended to write $\displaystyle \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3}{(x-5)^5(4-x)}\text.$
We have
$\displaystyle \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3}{(x-5)^5(4-x)}$
$\displaystyle =\lim_{x\to\infty}\frac{x^8(1-3/x)^8\left[x^3(2/x-1)^3\right]}{x^5(1-5/x)^5\left[x(4/x-1)\right]}$
$\displaystyle =\lim_{x\to\infty}\frac{x^{11}}{x^6}=\lim_{x\to\in fty}x^5=\infty$