Evaluating Limits

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January 16th 2011, 06:04 AM
Hockey_Guy14

2 Attachment(s)

Evaluating Limits

Evaluate the following limits:

a). Attachment 20456
I cant seem to figure out how to evaluate this and what the answer is, I thought it would be infinite or does not exist but thats not correct. Any help?

b)Attachment 20457
Cant seem to figure this one out either, I tried factoring out and dividing but its not working. Any help? Thanks alot (Nod)
January 16th 2011, 06:07 AM
Also sprach Zarathustra

a. Divide the numerator and denominator by x^2.

b. Divide the numerator and denominator by x^3.
January 16th 2011, 06:14 AM
HallsofIvy

In general, to find the limit, as x goes to infinity, of one polynomial over another, divide both numerator and denominator by the highest power of x that appears in either polynomial. That way each term will be either a constant or 1/x, to a power. And the limit, as x goes to infinity, of 1/x to any power is 0.

From that you can get a more specific rule for such a limit: if the numerator has higher power than the denominator, the limit does not exist. If the denominator has higher power than the numerator, the limit is 0. If both numerator and denominator have the same power, the limit is the ratio of the leading coefficients.
January 16th 2011, 06:48 AM
Prove It

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

$\displaystyle = \lim_{x\to \infty}\frac{1}{x}$

$\displaystyle = 0$ .
January 17th 2011, 09:36 AM
Hockey_Guy14

thanks for the help i got it now!
January 18th 2011, 08:46 AM
HallsofIvy

Quote:

Originally Posted by Prove It

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

$\displaystyle = \lim_{x\to \infty}\frac{1}{x}$

$\displaystyle = 0$ .

Yes, that's another way to do it!