1. ## L'Hopital's Rule question

I have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:

Find the limit of [ln(x)]/[ln(2x)] as x goes to infinity.

I tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?

2. Originally Posted by pianopiano
I have a problem using L'Hopital's rule and I can't seem to get the right answer. The problem is this:

Find the limit of [ln(x)]/[ln(2x)] as x goes to infinity.

I tried l'Hopital's rule, differentiating the top and the bottom, and then taking the limit of that expression. The derivative of the top is 1/x, and the derivative of the denominator is 1/4x (according to my calculations anyways). I find the limit of that expression, and I find that it is 4. Is that correct? However, I graph it on my calculator, and the limit seems to be 1. How do you get that?
Note that $\displaystyle \frac{\ln x}{\ln(2x)}=\frac{\ln x}{\ln 2+\ln x}$. Since $\displaystyle \lim_{x\to\infty}\frac{\ln x}{\ln(2x)}=\lim_{x\to\infty}\frac{\ln x}{\ln 2+\ln x}=\frac{\infty}{\infty}$, we can apply L'Hopitals rule.

So $\displaystyle \lim_{x\to\infty}\frac{\ln x}{\ln(2x)}=\lim_{x\to\infty}\frac{\ln x}{\ln 2+\ln x}=\lim_{x\to\infty}\frac{\displaystyle\frac{1}{x} }{\displaystyle\frac{1}{x}}=\lim_{x\to\infty}\frac {x}{x}=1$

Does this make sense?

3. Thanks, it does make sense!

However, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?

4. Originally Posted by pianopiano
Thanks, it does make sense!

However, why can't you differentiate the numerator and denominator and take the limit of the new expression and have that work as well since ln(x) and ln(2x) would each still go to infinity as x goes to infinity?
You could...but you would have to note that $\displaystyle \forall a\in\mathbb{R}\backslash\left\{0\right\},~\frac{\, d}{\,dx}\left[\ln\!\left(ax\right)\right]=\frac{1}{x}$

5. $\displaystyle \lim_{x\to\infty}\frac{\log x}{\log 2 + \log x}=\lim_{x\to\infty}\frac{1}{\underbrace{\frac{\lo g 2}{\log x}}_{\to 0}+1}=1$.

Or you must use L'H?

6. Originally Posted by Abu-Khalil
$\displaystyle \lim_{x\to\infty}\frac{\log x}{\log 2 + \log x}=\lim_{x\to\infty}\frac{1}{\underbrace{\frac{\lo g 2}{\log x}}_{\to 0}+1}=1$.

Or you must use L'H?

I was going to point that out.
There's no need for L'Hopital's Rule at all, just divide by the common term, log x.