Question

x-intercepts of: y=x^2*ln|x|+ex^2

Answer 1

\[y=x^2\ln|x|+ex^2\], yes that is the constant "e" times x, not the exponential e^x.

I end up with \[x^2(\ln|x|+e)=0 \implies x^2=0, \ln|x|=e^{-e}\]
\[\implies x=0, x=\pm e^{-e}\]

However, shouldn't the function not be defined at x=0, since ln(0) is undefined?
Wolfram gives me this: http://www.wolframalpha.com/input/?i=x-intercepts+x%5E2ln%7Cx%7C%2Bex%5E2
where one zero is (0,inderteminate)... what does this mean?

Answer 2

You have it right. x = 0 is NOT in the Domain.

What is the limit of the function as x approaches zero?

Answer 3

it would be 0?

Answer 4

Yes. The right limit is zero and the left limit is zero. Nevertheless, it is not continuous.