# limits problem..

• Mar 24th 2009, 03:59 AM
jvignacio
limits problem..
Hi guys, could i please get some help on this practice question.

Consider the function:

$
g(x) = \frac{5x^2}{x^2 + 2x - 3}
$
for $x \notin$ {-3,1}

find the following:

(a) $\lim_{x \to +\infty}$ $g(x)$

much appreciated
• Mar 24th 2009, 04:22 AM
Thomas154321
$g(x) = \frac{5x^2}{x^2 + 2x - 3}=5 - \frac{10x-15}{x^2 + 2x - 3}=5 - \frac{\frac{10}{x}-\frac{15}{x^2}}{1+\frac{2}{x}-\frac{3}{x^2}}$

Can you take it from here?
• Mar 24th 2009, 04:23 AM
HallsofIvy
Quote:

Originally Posted by jvignacio
Hi guys, could i please get some help on this practice question.

Consider the function:

$
g(x) = \frac{5x^2}{x^2 + 2x - 3}
$
for $x \notin$ {-3,1}

find the following:

(a) $\lim_{x \to +\infty}$ $g(x)$

much appreciated

"infinity" is hard to deal with but 0 is easy: divide both numerator and denominator by the highest power of x, $x^2$:
$\frac{5}{1+ \frac{2}{x}- \frac{3}{x^2}}$
Now, as x goes to infinity, the fractions with x in the denominator go to 0.
• Mar 24th 2009, 04:50 AM
jvignacio
Quote:

Originally Posted by Thomas154321
$g(x) = \frac{5x^2}{x^2 + 2x - 3}=5 - \frac{10x-15}{x^2 + 2x - 3}=5 - \frac{\frac{10}{x}-\frac{15}{x^2}}{1+\frac{2}{x}-\frac{3}{x^2}}$

Can you take it from here?

sorry im lost .. im not quite sure what you did to get from the first step to the second.
• Mar 24th 2009, 07:10 AM
Thomas154321
Quote:

Originally Posted by jvignacio
sorry im lost .. im not quite sure what you did to get from the first step to the second.

Algebraic long division. Or just say $5x^2 = (5x^2 + 10x -15) -10x + 15$. Then the numerator can be split up into 5 times the denominator with a remainder term. It's easier to do it like HallsofIvy suggests; I just prefer to take limits that tend to zero where possible with polynomial limits (for me at least it makes the working clearer).
• Mar 24th 2009, 01:59 PM
jvignacio
Quote:

Originally Posted by Thomas154321
Algebraic long division. Or just say $5x^2 = (5x^2 + 10x -15) -10x + 15$. Then the numerator can be split up into 5 times the denominator with a remainder term. It's easier to do it like HallsofIvy suggests; I just prefer to take limits that tend to zero where possible with polynomial limits (for me at least it makes the working clearer).

okay i know how to do Algebraic long division but i still cant seem to understand whats happening :(
• Mar 24th 2009, 02:52 PM
HallsofIvy
Quote:

Originally Posted by jvignacio
okay i know how to do Algebraic long division but i still cant seem to understand whats happening :(

What is $\lim_{x\rightarrow \infty} \frac{1}{x}$?

What is $\lim_{x\rightarrow \infty} \frac{5}{x^2}$?

Do you see the point?
• Mar 25th 2009, 05:27 AM
jvignacio
Quote:

Originally Posted by HallsofIvy
What is $\lim_{x\rightarrow \infty} \frac{1}{x}$?

What is $\lim_{x\rightarrow \infty} \frac{5}{x^2}$?

Do you see the point?

sorry for not understanding but what point are u trying to show me?
• Mar 25th 2009, 06:25 AM
Krizalid
Pay attention and read post #3.
• Mar 25th 2009, 07:32 AM
Pinkk
The general rule of thumb is to take the limit of the highest power of the numerator over the highest power of the denominator. So in your problem:

$\lim_{x \to +\infty}g(x)=\lim_{x \to +\infty} \frac{5x^{2}}{x^{2}} = \lim_{x \to +\infty} 5 = 5$. The above posts are proving to why that is true.

Quote:

Originally Posted by jvignacio
sorry im lost .. im not quite sure what you did to get from the first step to the second.

He kept dividing until he reached a point where you get:

$\lim_{x \to +\infty} g(x)= 5-\frac{0-0}{1+0-0} =5$

Remember:
If $f(x) = \frac{1}{x^{n}}$ where $n > 0$, then:

$\lim_{x \to +\infty} f(x) = 0$