Rewrite the limit as
[tex]\displaystyle\lim_{x\to0}x^2\log x^2=\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}[/tex]
Then both numerator and denominator approach infinity (with different signs, but that's not important). Applying L'Hopital's rule, we get
[tex]\displaystyle\lim_{x\to0}\frac{\log x^2}{\frac1{x^2}}=\lim_{x\to0}\frac{\frac2x}{-\frac2{x^3}}=\lim_{x\to0}-x^2=\boxed{0}[/tex]
