Another limit task, x over ln x, L'Hôpital won't do it?

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

This was wrong, I got a big red wrong! Why doesn't L'Hôpital work on this one? The problem is that $\ln$ is not defined for 0. It needs to be rewritten?

(Thanks to everyone helping me out with my homework, due anxiety I'm not able attend the class workshops.)

Edit: I did get the correct limit ($0$), but that was a coincidence.

Solution 1:

L'Hopital's rule is best forgotten about. Of course as $x\to0^+$, $\log x\to-\infty$ and so $1/\log x\to0$. A fortiori $x/\log x\to 0$.

A better problem is to find $\lim_{x\to0^+}x\log x$.

Solution 2:

The limit, as written, does not exist because the function is not defined on an open neighborhood of the limit point: the function is not defined on $(-\infty,0)$, so you cannot take the limit as $x\to 0$. That would be a full answer.

If, however, the limit is taken only from the right, then the other answers you have received are correct: the limit is not an indeterminate form, it is equal to $0$. But you cannot get that by 'evaluating', because $\ln(x)$ is not defined at $0$.

Solution 3:

$\frac{0}{\infty}$ is not indeterminate form, is zero, then $\lim\limits_{ x \rightarrow 0^+ }\frac{x}{lnx}=0$ but is a right limit.

But L'Hôpital's rule applies..

Solution 4:

As others have mentioned, you don't need L'Hospital's rule for this simple example. However, in fact, there is a more general form of L'Hôpital's rule that requires only that the denominator $\to\infty\:.$ This handles your case and many others so it is well-worth knowing. For an exposition see the papers below and see the theorem in Rudin's PoMA excerpted in my answer here.

A. E. Taylor, $\ $ L'Hospital's Rule
Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24.
http://www.jstor.org/stable/2307183

A. M. Ostrowski, $\ $ Note on the Bernoulli-L'Hospital Rule
Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242.
http://www.jstor.org/stable/2318210