hi do anyone knows what ∞ - ln∞ gives?
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$\displaystyle \infty- ln\infty$ 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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$\displaystyle \displaystyle f(x) = x - \frac{3}{2}\ln{(x^2+2)}$
$\displaystyle \displaystyle f'(x) = 1 - \frac{3x}{x^2 + 2} \to 1$ as $\displaystyle x \to \infty$.
Since the derivative tends to a positive number, the function will continue to increase without bound.
So $\displaystyle \displaystyle \lim_{x \to \infty} f(x) = \infty$.
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: .$\displaystyle \displaystyle \lim_{x\to\infty}\,\ln\left(\dfrac{e^x}{x^3}\right ) \;=\;\infty$