l'Hopital's rule

**crossbone** · February 22nd 2011, 02:37 PM

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

**skeeter** · February 22nd 2011, 04:20 PM

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

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

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

the limit DNE

**crossbone** · February 22nd 2011, 04:32 PM

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

**skeeter** · February 22nd 2011, 06:18 PM

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 (x*log(x))/(log(1& #43;2x)) - Wolfram|Alpha

**crossbone** · February 22nd 2011, 07:35 PM

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

**Prove It** · February 22nd 2011, 07:43 PM

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

**crossbone** · February 22nd 2011, 07:57 PM

I knew that but this is what I got:
$ \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 \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
$ k=a\cdot(a-1)\cdot(a-2)\cdot (a-3).....(a-n+1) $

Is this right?

**Prove It** · February 22nd 2011, 08:49 PM

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

You need to use the chain rule...

**AllanCuz** · February 22nd 2011, 09:27 PM

Originally Posted by skeeter

I would say not ... the original limit does not exist.

limit as x to 0 (x*log(x))/(log(1& #43;2x)) - Wolfram|Alpha

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?

**FernandoRevilla** · February 22nd 2011, 10:01 PM

Originally Posted by crossbone

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

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

Then reapply l'Hopital's rule again:
$ \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:

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

Fernando Revilla

**crossbone** · February 22nd 2011, 11:11 PM

Originally Posted by Prove It

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

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 \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}} $

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