limits of horizontal asymptotes

**used9090** · February 21st 2009, 05:04 PM

How do you figure out the:

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

**JoshHJ** · February 21st 2009, 05:33 PM

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 $x^6$ , and the highest term in the numerator would be $x^{11}$ . This way you can find out that there will be a $x^5$ expression left meaning the limit as x approaches infinity is infinity.

**Reckoner** · February 21st 2009, 05:38 PM

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 $\lim_{x\to\infty}\frac{(x-3)^8(2-x)^3(4-x)}{(x-5)^5},$ but I am guessing that you intended to write $\lim_{x\to\infty}\frac{(x-3)^8(2-x)^3}{(x-5)^5(4-x)}\text.$

We have

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

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

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

**used9090** · February 21st 2009, 07:39 PM

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

**JoshHJ** · February 21st 2009, 07:42 PM

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.

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