Solving $\lim_{x\rightarrow0+} x^2 \ln x$ with L'Hoptial's Rule

[DeadWolfe]

I was asked to find

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

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?

 

[StatusX]

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

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

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

$$-x^2/2$$

$$0$$

 

[Zlex]

I was asked to find

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

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?

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

 

[DeadWolfe]

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?

 

[addle_brains]

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.

 

[Cyosis]

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

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

 

[HallsofIvy]

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 e^x is e^x. 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.

 

[addle_brains]

Thanks guys. That helps a lot. I still don't know every little trick of derivatives and such.