hi do anyone knows what ∞ - ln∞ gives?

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- Oct 31st 2010, 03:36 AMfirebrendanswer of ∞ - ln∞
hi do anyone knows what ∞ - ln∞ gives?

[IMG]file:///C:/Users/Arwin/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG][IMG]file:///C:/Users/Arwin/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG] - Oct 31st 2010, 03:41 AMScurmicurv
It gives infinity. ln∞ does tend to infinity, but not very fast.

Keep in mind though, that infinity isn't like any ordinary number. ∞+∞ = ∞, for example. - Oct 31st 2010, 03:45 AMmr fantastic
- Oct 31st 2010, 04:21 AMHallsofIvy
$\displaystyle \infty- ln\infty$ doesn't give anything. "Infinity" is not a real number and ordinary functions such as "logarithm" and "subtraction" do not apply to it. There are extended number systems in which "infinity" is defined as a number but they still do not apply "logarithm" or "subtraction" in the usual sense.

Something of the form lim [f(x)- ln(g(x))] where $\displaystyle lim f= \infty$ and $\displaystyle lim g= \infty$ may exist but the limit will depend upon exactly**how**f and g "go to infinity".

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- Oct 31st 2010, 05:38 AMfirebrend
The question is:

given: f(x)= x-3/2 ln〖(x^2+2) 〗

Using mathematical arguments, investigate what happens to f(x) as x -> ∞ - Oct 31st 2010, 06:02 AMProve It
$\displaystyle \displaystyle f(x) = x - \frac{3}{2}\ln{(x^2+2)}$

$\displaystyle \displaystyle f'(x) = 1 - \frac{3x}{x^2 + 2} \to 1$ as $\displaystyle x \to \infty$.

Since the derivative tends to a positive number, the function will continue to increase without bound.

So $\displaystyle \displaystyle \lim_{x \to \infty} f(x) = \infty$. - Oct 31st 2010, 07:31 AMSoroban
Hello, firebrend!

Here is a reallysolution . . .*sloppy*

Quote:

$\displaystyle \text{Given: }\:f(x) \:=\:x - \tfrac{3}{2}\ln(x^2+2) $

$\displaystyle \displaystyle \text{Investigate: }\;\lim_{x\to\infty}f(x)$

We have: .$\displaystyle f(x) \:=\:x - \ln(x^2+2)^{\frac{3}{2}}$

For very large $\displaystyle \,x,\:f(x) \:\approx\:x - \ln(x^2)^{\frac{3}{2}} \;=\;x - \ln(x^3) $

Then: .$\displaystyle f(x) \;\approx\;\ln(e^x) - \ln(x^3) \;=\;\ln\left(\dfrac{e^x}{x^3}\right) $

It can be shown that: .$\displaystyle \displaystyle \lim_{x\to\infty}\,\ln\left(\dfrac{e^x}{x^3}\right ) \;=\;\infty$