1> lim ((3x^3)-2x) / ((x^2)-1)

x->infinite

2> lim 1/ |x-3|

x->3-

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- May 10th 2006, 05:04 AMbobby77please very urgent
1> lim ((3x^3)-2x) / ((x^2)-1)

x->infinite

2> lim 1/ |x-3|

x->3- - May 10th 2006, 09:23 AMCaptainBlackQuote:

Originally Posted by**bobby77**

$

So as $\displaystyle x\to \infty$ the top $\displaystyle \to \infty$ and the

bottom $\displaystyle \to 1$, so:

$\displaystyle

\lim_{x \to \infty} \frac{3x^3-2x}{x^2-1}=\infty

$

which is short hand for it diverges.

RonL - May 10th 2006, 09:23 AMJameson
$\displaystyle \lim_{x\to\infty}\frac{3x^3-2x}{x^2-1}$

Look at the highest power of the numerator versus the denominator. The numerator has a degree of 3 while the denominator has a degree of two. You should see that as as grows larger and larger, the top far outgrows the bottom and this limit is infinity.

$\displaystyle \lim_{x\to{3-}}\frac{1}{|x-3|}$

You should first realize that this is a form of the rational graph $\displaystyle \frac{1}{x}$ shifted to the right 3 units, and all negative values are flipped about the x-axis. Here's how I would do this problem. Test the value -2.9. How big is that number? Is it positive or negative? Now test -2.99, then -2.9999. See the trend?