# Limit invloving lnx

1. Nov 17, 2004

$$\lim_{x\rightarrow0+} x^2 \ln x$$

Using L'Hoptial's rule.

I guess that I have to divide by one of these things to the power of negative one, and then take the derivative of top and bottom, but this results in either dividing by zero if I put x^2 on the bottom, or not losing ln x if I put it on the bottom, because of the chain rule.

I really am doubting that L'Hopital is the way to go about this, but that's what I was told on a test today.

Any ideas?

2. Nov 17, 2004

### StatusX

You aren't dividing by 0 by putting x^2 on the bottom. Remember, this is a limit, the values aren't fixed.

$$\frac{ln(x)}{1/x^2}$$

$$\frac{1/x}{-2/x^3}$$

$$-x^2/2$$

$$0$$

3. Nov 17, 2004

### Zlex

$$\lim_{x\rightarrow0+} \frac{x^3 \ln x}{x}$$

4. Nov 17, 2004

StatusX - thank you, I tihnk that works

Zlex - do you see how that method will not eliminate the ln x because of the product rule?

5. May 7, 2009

Hey everyone. I know this question is old but I'm trying to solve exactly the same problem. Could anyone explain it to me a little better? I'm having trouble seeing how StatusX got the answer.

I understand L'Hopital's rule but how the derivatives of f(g) and f(h) were found is confusing me.

Last edited: May 7, 2009
6. May 7, 2009

### Cyosis

The derivatives of f(g) and f(h)? Do you mean f(x) and g(x)? Either way they did the following:

$$x^2 \ln x=\frac{\ln x}{x^{-2}} \Rightarrow f(x)=\ln x, g(x)=x^{-2} \Rightarrow \lim_{x \to 0+}x^2 \ln x=\lim_{x \to 0+} \frac{f(x)}{g(x)}=\lim_{x \to 0+} \frac{f'(x)}{g'(x)}=\lim_{x \to 0+} \frac{\frac{1}{x}}{-2 x^{-3}}=\lim_{x \to 0+} -2 x^2$$.

7. May 7, 2009

### HallsofIvy

Staff Emeritus
The numerator, in Statusx's form, is ln(x), and the denominator is 1/x2= x-2.

The derivative of ln(x) is a "standard form", derived from the fact that the derivative of ex is ex. d(ln(x))/dx= 1/x.

The derivative of x-2 is, of course, from the power rule: (-2)x-2-1= -2x-3= -2/x3.

8. May 7, 2009