# Math Help - l'Hopital applies or not?

1. ## l'Hopital applies or not?

Kinda weird how little I could glean about this one on the webernet. Oh, there were TONS of other examples, but this one is, uhh, too simple? Maybe?

$\lim_{x\to\0+}\frac{ln x}{x}$

I know that the answer is $-\infty$. I know multiple ways to figure out the answer. My question to you is if I can use l'Hopital's rule on it?

If you use it without factoring first, you get $\frac{\frac{1}{x}}{1}$ OR $\frac{1}{x}$ which will tend to $\infty$ as $x \to\0$. But if you factor first, you get $\lim_{x\to\0}\frac{1}{x}$ $\cdot$ $\lim_{x\to\0} ln x$. Now from here you can just say it's $\infty\cdot-\infty=-\infty$ and be done with it, but I don't think that holds up to a math teacher (does it?) so IF I continue and get derivatives then it becomes $\lim_{x\to\0}-\frac{1}{x^2}$ $\cdot$ $\lim_{x\to\0}\frac{1}{x}$ which can be written as $\lim_{x\to\0}-\frac{1}{x^3}=-\infty$.

Intuitively this SEEMS to make sense, but am I missing something? I don't want to demonstrate this only to lose points for not seeing that l'hopital's rule doesn't apply. Can someone show me the light?

Thanks!

2. ## Re: l'Hopital applies or not?

Originally Posted by rortiz81
Kinda weird how little I could glean about this one on the webernet. Oh, there were TONS of other examples, but this one is, uhh, too simple? Maybe?

$\lim_{x\to\0+}\frac{ln x}{x}$

I know that the answer is $-\infty$. I know multiple ways to figure out the answer. My question to you is if I can use l'Hopital's rule on it?

If you use it without factoring first, you get $\frac{\frac{1}{x}}{1}$ OR $\frac{1}{x}$ which will tend to $\infty$ as $x \to\0$. But if you factor first, you get $\lim_{x\to\0}\frac{1}{x}$ $\cdot$ $\lim_{x\to\0} ln x$. Now from here you can just say it's $\infty\cdot-\infty=-\infty$ and be done with it, but I don't think that holds up to a math teacher (does it?) so IF I continue and get derivatives then it becomes $\lim_{x\to\0}-\frac{1}{x^2}$ $\cdot$ $\lim_{x\to\0}\frac{1}{x}$ which can be written as $\lim_{x\to\0}-\frac{1}{x^3}=-\infty$.

Intuitively this SEEMS to make sense, but am I missing something? I don't want to demonstrate this only to lose points for not seeing that l'hopital's rule doesn't apply. Can someone show me the light?

Thanks!
Infinity is not a number, so you can't expect numerical rules to work with it. Use L'Hospital.

3. ## Re: l'Hopital applies or not?

No, L'Hopital's rule does NOT apply because the numerator and denominator are not both going to the same thing. L'Hopital's rule only applies if numerator an denominator both go to 0, or both to positive infinity, or negative infinity.

Here, it should be trivial to see that for x very close to 0, from above, the numerator is a very large, negative, number while the denominator is a very small positive number. The quotient is a very large negative number so the limit is negative infinity.

4. ## Re: l'Hopital applies or not?

Originally Posted by HallsofIvy
No, L'Hopital's rule does NOT apply because the numerator and denominator are not both going to the same thing. L'Hopital's rule only applies if numerator an denominator both go to 0, or both to positive infinity, or negative infinity.

Here, it should be trivial to see that for x very close to 0, from above, the numerator is a very large, negative, number while the denominator is a very small positive number. The quotient is a very large negative number so the limit is negative infinity.
I made a mistake, I assumed a transformation could be applied to get 0/0.

5. ## Re: l'Hopital applies or not?

Thanks guys! But you're juxtaposing answers just confuse me even more. Is what I did wrong, even figuring it wasn't L'Hopital rules?

6. ## Re: l'Hopital applies or not?

FYI I know reasoning or just making a data table can tell me it's neg. infinity, but I guess my ultimate question is-Is reasoning the ONLY way to evaluate this limit?