l'Hopital's rule

[crossbone]

Find $\displaystyle \lim_{x \to 0^{+}}\frac{x \log{x}}{\log{(1+2x)}}$

Can I do this:
$\displaystyle \lim_{x \to 0^{+}}\frac{x \log{x}}{\log{(1+2x)}}= \frac{\lim_{x \to 0^{+}} x \log{x}}{\lim_{x \to 0^{+}} \log{(1+2x)}}=\frac{\lim_{x \to 0^{+}} \frac{\log{x}}{\frac{1}{x}}}{\lim_{x \to 0^{+}} \log{(1+2x)}}=\frac{\lim_{x \to 0^{+}} -x}{\lim_{x \to 0^{+}} \log{(1+2x)}}$

where l'Hopital's rule is applied to the numerator only. Then reapply l'Hopital's rule again:

$\displaystyle \lim_{x \to 0^+} \frac{-x}{\log (1+2x)}=\lim_{x \to 0^+} \frac {-1}{\frac{2}{1+2x}}=-\frac{1}{2}$

Lemme know if my divide and conquer approach with l'Hopital's rule is legit.

 

[tkhunny]

Sort of, but please recall that the rule does NOT apply unless you have an indeterminate form.

It's pretty obvious that the limit of the denominator is zero (0).

Have you PROVEN that the limit of the numerator is zero (0)?

Let's see.

Try x = 1: $\displaystyle \frac{1\cdot log(1)}{log(1+2\cdot 1)} = 0$

Try x = 1/5: $\displaystyle \frac{\frac{1}{5}log(1/5)}{log(1+2*(1/5))} = \frac{-log(5)}{5\cdot log(7/5)} = -0.957$

Whoops. Rather shoots holes in the -1/2 theory. Better give it another go.

 

[crossbone]

Thanks for the reply. What I got is $\displaystyle 0\cdot \infty$ for the numerator which is an indeterminate form for l'Hopital's rule and thought it was ok to apply the rule there first then reapply it when taking the denominator into consideration. I guess this is wrong.

 

[tkhunny]

No, that's ok, but once you reconcile that limit, the whole thing still has to be an indeterminate form. This you did. Very good.

Your error was when you started with $\displaystyle \frac{-x}{log(1+2x)}$. That's a derivative in the numerator and the original in the denominator. After proving that the original numerator's limit is zero, start the whole thing over from the original expression. Then you will see it. Playing with just the numerator is ONLY to establish that the original is an appropriate indeterminate form. Start over from the beginning after that.

 

[crossbone]

Thanks! I got it now. I got another problem though. Just checking if this is right:
$\displaystyle \displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}=\lim_{x \to 0} \frac{a|\log{x}|^{a-1}}{\frac{-1}{x}}$
keep differentiating until a-n=b<0:
$\displaystyle \displaystyle \lim_{x \to 0} {(-)}^n k \frac{|\log{x}|^b}{\frac{1}{x}}=\lim_{x \to 0} {(-)}^n k x|\log{x}|^b=0$
where
$\displaystyle k=a\cdot(a-1)\cdot(a-2)\cdot (a-3).....(a-n+1)$

 

[DrSteve]

$\displaystyle \log x$ is undefined for $\displaystyle x<0$. Thus this limit does not exist (this is immediate). Did you perhaps mean to be taking the limit from the right?

Also, $\displaystyle |\log x|=-\log x$ near $\displaystyle 0$. I'm not sure if this is clear in your computation.

 

[lookagain]

Thanks! I got it now. I got another problem though. Just checking if this is right:

Suppose $\displaystyle a$ is a positive integer.


Evaluate the following if the limit exists:

$\displaystyle \displaystyle \lim_{x \to 0}\frac{(\log{|x|})^a}{\frac{1}{x}}$


This verion would make more sense to me, because "from the right" wouldn't have to be stated.