- #1
DeadWolfe
- 457
- 1
I was asked to find
[tex]\lim_{x\rightarrow0+} x^2 \ln x [/tex]
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?
[tex]\lim_{x\rightarrow0+} x^2 \ln x [/tex]
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?