1. ## l'Hopital's rule

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.

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

$\displaystyle \displaystyle\lim_{x \to 0^+} \frac{1+2x}{2} \cdot \lim_{x \to 0^+} \log{x}$

$\displaystyle \displaystyle \frac{1}{2} \cdot \lim_{x \to 0^+} \log{x}$

the limit DNE

3. so divide and conquer with l'Hopital and reapplying it is not legal?

4. Originally Posted by crossbone
so divide and conquer with l'Hopital and reapplying it is not legal?
I would say not ... the original limit does not exist.

limit as x to 0 &#40;x&#42;log&#40;x&#41;&#41;&#47;&#40;log&#40;1& #43;2x&#41;&#41; - Wolfram|Alpha

5. Here's another problem: $\displaystyle \displaystyle \lim_{x \to 0} x |{\log x|^a$

6. Rewrite it as $\displaystyle \displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}$, which is now of the form $\displaystyle \displaystyle \frac{\infty}{\infty}$ so you can now use L'Hospital's Rule...

7. I knew that but this is what I got:
$\displaystyle \displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}=\lim_{x \to 0} \frac{a|\log{x}|^{a-1}}{\frac{-1}{x}}$
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
$\displaystyle k=a\cdot(a-1)\cdot(a-2)\cdot (a-3).....(a-n+1)$

Is this right?

8. The derivative of $\displaystyle \displaystyle |\log{x}|^a$ is not $\displaystyle \displaystyle a|\log{x}|^{a-1}$.

You need to use the chain rule...

9. Originally Posted by skeeter
What exactly does this approach violate? I'm looking at the rules for L`Hopitals and I don't quite see where this goes wrong, but it obviously does violate something: http://bradley.bradley.edu/~delgado/121/Lhopital.pdf

Any ideas?

10. Originally Posted by 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.

Right. With this, you prove

$\displaystyle \displaystyle \lim_{x \to 0^{+}}\dfrac{f(x)}{g(x)}=\ldots=\dfrac{0}{0}$

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.

But now, you find:

$\displaystyle \displaystyle \lim_{x \to 0^{+}}\dfrac{f''(x)}{g'(x)}$

Fernando Revilla

11. Originally Posted by Prove It
The derivative of $\displaystyle \displaystyle |\log{x}|^a$ is not $\displaystyle \displaystyle a|\log{x}|^{a-1}$.

You need to use the chain rule...
I did. it appears I didn't cos I simplified it after I used l'Hopital's rule:

$\displaystyle \displaystyle \lim_{x \to 0}\frac{|\log{x}|^a}{\frac{1}{x}}=\lim_{x \to 0} \frac{\frac{a}{x}|\log{x}|^{a-1}}{\frac{-1}{x^2}}=\lim_{x \to 0}\frac{a|\log{x}|^{a-1}}{\frac{-1}{x}}$