1. ## please very urgent

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

2> lim 1/ |x-3|
x->3-

2. Originally Posted by bobby77
1> lim ((3x^3)-2x) / ((x^2)-1)
x->infinite
$\displaystyle \lim_{x \to \infty} \frac{3x^3-2x}{x^2-1}=\lim_{x \to \infty} \frac{3x-2/x}{1-1/x^2}$

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

3. $\displaystyle \lim_{x\to\infty}\frac{3x^3-2x}{x^2-1}$

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?