1. ## problematic limit

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.

2. As x tends to zero, the given expression becomes an indeterminate of the form $0.\infty$

So convert it into an indeterminate form which allows application of L'Hospital rule .

Like this : $f(x) = x({\ln(x))}^{23}$
$= ({\ln(x))}^{23}/(1/x)$

Now consider the numerator, $({\ln(x))}^{23}$ and the denominator $(1/x)$

Both terms become infinite as x tends to zero, so L'Hospital's rule is applicable.

3. In fact you can apply L'hopital:

$x\cdot\ln(x)^{23} = \frac{\ln(x)^{23}}{\frac{1}{x}} = [-\frac{\infty}{\infty}]$ (as x --> 0)

by repeatedly applying L'hopital you should find: $\lim_{x\to 0}x\cdot \ln(x)^{23} = \lim_{x\to 0}-23! x = 0$

4. Hello, waytogo!

Ah, Dinkydoe beat me to it . . .
Well, I'm not going to delete all this!

$\lim_{x\to0}x\!\cdot\!(\ln x)^{23}$

As given, the limit is: . $0\cdot(-\infty)$

We have: . $\frac{(\ln x)^{23}}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}$ . .
We can apply L'Hoptal's Rule

Apply L'Hopital: . $\frac{23\cdot (\ln x)^{22}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot(\ln x)^{22}}{x^{-1}}\;\rightarrow\;\frac{\text{-}\infty}{\infty}$ . .
We can apply L'Hoptal's Rule.

Apply L'Hopital: . $-\frac{23\cdot22\cdot(\ln x)^{21}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;\frac{23\cdot22\cdot(\ln x)^{21}}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}$

Apply L'Hopital: . $\frac{23\cdot22\cdot21\cdot(\ln x)^{20}\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot22\cdot21\cdot(\ln x)^{20}}{x^{-1}} \;\rightarrow\; \frac{\text{-}\infty}{\infty}$

. . . . . $\vdots$

Apply L'Hopital: . $\frac{23\cdot22\cdot21\cdots 3\cdot(\ln x)\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23\cdot22\cdot21\cdots3\cdot(\ln x)^2}{x^{-1}} \;\rightarrow\;\frac{\text{-}\infty}{\infty}$

Apply l"Hopital: . $\frac{23\cdot22\cdot21\cdots2\cdot(\ln x)\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;\frac{23!(\ln x)}{x^{-1}} \;\rightarrow\; \frac{\text{-}\infty}{\infty}$

Apply L'Hopital: . $\frac{23!\left(\frac{1}{x}\right)}{-x^{-2}} \;=\;-\frac{23!}{x^{-1}} \;=\;-23!x$

Therefore: . $\lim_{x\to0}\,(-23!x) \;=\;0$