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hi!
i have a problem in finding limit for this funciton f (x) = x*(lnx)^23,where x --> 0. i thought i will be able to use Lopital's rule, but cannot transformate funcition to uncertainity in order to use Lopital's rule. Do anyone have ideas? And I also think that there should be way to find limit without Lopital's rule, using formulas like e^x-1~x.
But anyway, i have no ideas for the moment. Can anybody help?
thanks.
As x tends to zero, the given expression becomes an indeterminate of the form http://latex.codecogs.com/gif.latex?0.\infty
So convert it into an indeterminate form which allows application of L'Hospital rule .
Like this : http://latex.codecogs.com/gif.latex?...{\ln(x))}^{23}
http://latex.codecogs.com/gif.latex?...))}^{23}/(1/x)
Now consider the numerator, http://latex.codecogs.com/gif.latex?({\ln(x))}^{23} and the denominator http://latex.codecogs.com/gif.latex?(1/x)
Both terms become infinite as x tends to zero, so L'Hospital's rule is applicable.
In fact you can apply L'hopital:
$\displaystyle x\cdot\ln(x)^{23} = \frac{\ln(x)^{23}}{\frac{1}{x}} = [-\frac{\infty}{\infty}]$ (as x --> 0)
by repeatedly applying L'hopital you should find: $\displaystyle \lim_{x\to 0}x\cdot \ln(x)^{23} = \lim_{x\to 0}-23! x = 0$
Hello, waytogo!
Ah, Dinkydoe beat me to it . . .
Well, I'm not going to delete all this!
Quote:
$\displaystyle \lim_{x\to0}x\!\cdot\!(\ln x)^{23}$
As given, the limit is: .$\displaystyle 0\cdot(-\infty)$
We have: .$\displaystyle \frac{(\ln x)^{23}}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}$ . . We can apply L'Hoptal's Rule
Apply L'Hopital: .$\displaystyle \frac{23\cdot (\ln x)^{22}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot(\ln x)^{22}}{x^{-1}}\;\rightarrow\;\frac{\text{-}\infty}{\infty}$ . . We can apply L'Hoptal's Rule.
Apply L'Hopital: .$\displaystyle -\frac{23\cdot22\cdot(\ln x)^{21}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;\frac{23\cdot22\cdot(\ln x)^{21}}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}$
Apply L'Hopital: .$\displaystyle \frac{23\cdot22\cdot21\cdot(\ln x)^{20}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot22\cdot21\cdot(\ln x)^{20}}{x^{-1}} \;\rightarrow\; \frac{\text{-}\infty}{\infty} $
. . . . . $\displaystyle \vdots$
Apply L'Hopital: .$\displaystyle \frac{23\cdot22\cdot21\cdots 3\cdot(\ln x)\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot22\cdot21\cdots3\cdot(\ln x)^2}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty} $
Apply l"Hopital: .$\displaystyle \frac{23\cdot22\cdot21\cdots2\cdot(\ln x)\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;\frac{23!(\ln x)}{x^{-1}} \;\rightarrow\; \frac{\text{-}\infty}{\infty}$
Apply L'Hopital: .$\displaystyle \frac{23!\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23!}{x^{-1}} \;=\;-23!x$
Therefore: .$\displaystyle \lim_{x\to0}\,(-23!x) \;=\;0$
