1. ## answer of ∞ - ln∞

hi do anyone knows what ∞ - ln∞ gives?

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2. 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.

3. Originally Posted by firebrend
hi do anyone knows what ∞ - ln∞ gives?

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Please post the actual question, exactly as it's written.

4. Originally Posted by firebrend
hi do anyone knows what ∞ - ln∞ gives?
$\displaystyle \infty- ln\infty$ 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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5. The question is:
given: f(x)= x-3/2 ln⁡〖(x^2+2) 〗

Using mathematical arguments, investigate what happens to f(x) as x -> ∞

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

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

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

So $\displaystyle \displaystyle \lim_{x \to \infty} f(x) = \infty$.

7. Hello, firebrend!

Here is a really sloppy solution . . .

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

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

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