# Urgent Help! with Limits

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• August 31st 2006, 05:33 AM
afn2
Urgent Help! with Limits
I need help solving what I think is a pretty simple limits questions, but I just can't see where to start.

I need to find:
the lim as x goes to infinity of (3x^2+2x+2)/(4x^2+2x+7)
and the lim as x goes to zero of the same function

Please help!
• August 31st 2006, 06:10 AM
galactus
When taking the limit of a rational expression, if x approaches +/- infinity, you can disregard all but the coefficients of the highest powers.

Therefore you have,

$\lim_{x\to\infty}\frac{3x^{2}}{4x^{2}}$

Now, you can see what it is?.

As x approaches 0, look close, can you see it?.
• August 31st 2006, 06:39 AM
ThePerfectHacker
Quote:

Originally Posted by afn2

I need to find:
the lim as x goes to infinity of (3x^2+2x+2)/(4x^2+2x+7)
and the lim as x goes to zero of the same function

Divide the numerator and denominator by $x^2$
$\lim_{x\to\infty}\frac{3+\frac{2}{x}+\frac{2}{x^2} }{4+\frac{2}{x}+\frac{7}{x^2}}$
The numerator is,
$\lim_{x\to\infty} 3+\frac{2}{x}+\frac{2}{x^2}=3$
The denominaot is,
$\lim_{x\to\infty}4+\frac{2}{x}+\frac{7}{x^2}=4$
Since both limits exists thus the value of the limit is 3/4
• August 31st 2006, 06:57 AM
topsquark
Quote:

Originally Posted by afn2
I need help solving what I think is a pretty simple limits questions, but I just can't see where to start.

I need to find:
the lim as x goes to infinity of (3x^2+2x+2)/(4x^2+2x+7)
and the lim as x goes to zero of the same function

Please help!

The limit as x goes to 0 should be simple. Both the numerator and denominator are continous functions on all real numbers (ie. the domain in both cases is all real numbers) and the denominator is not 0 for x = 0, so effectively all you need to do is substitute 0 for x in the expression.

-Dan
• August 31st 2006, 07:50 AM
afn2
Thanks a bunch, I knew it was pretty simple, but I've always struggled with limits, even though I completely get derivatives and integrals. I just needed a push in the right direction. Thanks guys!