1. limits of horizontal asymptotes

How do you figure out the:

lim X-> OO (x-3)^8 (2-X)^3/ (X-5)^5 (4-X)

2. Originally Posted by used9090
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.

3. Originally Posted by used9090
How do you figure out the:

lim X-> OO (x-3)^8 (2-X)^3/ (X-5)^5 (4-X)
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

$\displaystyle \lim_{x\to\infty}\frac{(x-3)^8(2-x)^3}{(x-5)^5(4-x)}$

$\displaystyle =\lim_{x\to\infty}\frac{x^8(1-3/x)^8\left[x^3(2/x-1)^3\right]}{x^5(1-5/x)^5\left[x(4/x-1)\right]}$

$\displaystyle =\lim_{x\to\infty}\frac{x^{11}}{x^6}=\lim_{x\to\in fty}x^5=\infty$

4. My teachers solution guide listed the answer to that specific problem as
-OO though. Is there an error perhaps in my teachers arithmetic?

5. Well if the problem is the limit as that function approaches infinity, then yes. But if the problem is the limit as that function approaches negative infinity, then your teacher is correct.