# Thread: Finding limit at infinity

1. ## Finding limit at infinity

Find the limit at infinity.

g(z) = 4z^6 - 7z^3 / (z^2 -4)^3

2. Expand the denominator and divide it out through the numerator, the form may then be easier to work with.

3. I expanded the denominator and jsut got one long exquation...
Not seeing how this helps...

Please show and explain

4. Not sure if this helps and plus I am also really new to these forums, but I remember my teacher taught me this:

Look at the highest power in the numerator and in the denominator.

1) If they both have the same highest power, the ratio between the numerator and the denominator is the limit.

2)If the denominator has a higher power, the limit goes to zero

3)If the numerator has a higher power, the limit goes to infinity

ex1) $5x^2 / x^2$ limit goes to 5

ex2) $5x / x^2$ limit goes to zero

ex3) $5x^2 / x$ limit goes to infinity

So for your problem, try to compare the highest powers and see what happens. I hope this may be of some help to you and I hope I remembered right

5. So the limit is 4...Right?

6. Yes, usually when I did these kinds of problems and found an answer I would check it with this approach:

Plug in a very large value into the equation (such as 500,000 or something), and see what happens. Make sure that you just don't go about doing this to find the answer, because you need to understand how that answer came to be. For this problem, plugging in a large value such as 500,000 would lead to the limit, 4.

7. Originally Posted by jzellt
I expanded the denominator and jsut got one long exquation...
Not seeing how this helps...

Please show and explain
$\frac{4z^6-7z^3}{z^6-12z^4+48z^2-64}$

now divide both numerator and denominator by $z^6$

$\frac{\frac{4z^6}{z^6}-\frac{7z^3}{z^6}}{\frac{z^6}{z^6}-\frac{12z^4}{z^6}+\frac{48z^2}{z^6}-\frac{64}{z^6}}$

$\frac{4-\frac{7}{z^2}}{1-\frac{12}{z^2}+\frac{48}{z^4}-\frac{64}{z^6}}$

So the limit will be 4.