Infinite limits

**plstevens** · January 31st 2008, 04:22 PM

Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)

how do i solve this problem.

**WWTL@WHL** · January 31st 2008, 04:33 PM

Hello

1)Expand the bottom out.
2)Divide both the top and bottom of the fraction by $u^4$

And you should arrive at:

$\lim_{u \to \infty} = \frac{3 + \frac{3}{u^4}}{2 - \frac{17}{u^2} + \frac{8}{u^4}}$

You know that $\lim_{u \to \infty} \frac{1}{u} = 0$ right? So anything like $\lim_{u \to \infty} \frac{3}{u^4}$ you can express as $3 \lim_{u \to \infty} \frac{1}{u^4}$ or even $3 \lim_{u \to \infty} \frac{1}{u} . \lim_{u \to \infty} \frac{1}{u}. \lim_{u \to \infty} \frac{1}{u} . \lim_{u \to \infty} \frac{1}{u} = 0$

So the limit in your question is $\frac{3 + 0}{2 - 0 - 0}$

= $\frac{3}{2}$

**topsquark** · January 31st 2008, 04:36 PM

Originally Posted by plstevens

Find the limit. (If you need to use - or , enter -INFINITY or INFINITY.)

how do i solve this problem.

Did you mean
$\lim_{u \to \infty} \frac{3u^4 + 3}{(u^2 - 8)(2u^2 - 1)}$

There are a variety of ways. I think the clearest at this stage of the game would be to expand the denominator.
$\lim_{u \to \infty} \frac{3u^4 + 3}{(u^2 - 8)(2u^2 - 1)}$

$= \lim_{u \to \infty} \frac{3u^4 + 3}{2u^4 - 17u^2 + 8}$

The next step is standard: Divide the numerator and denominator by the highest power appearing in the problem, in this case $u^4$ :
$= \lim_{u \to \infty} \frac{3 + \frac{3}{u^4}}{2 - \frac{17}{u^2} + \frac{8}{u^4}}$

Now take your limit.

-Dan

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