Question:
$$\lim_{{x,y}\to 0^+} \frac{x^{y^{x}}-x^{y^{2}}}{y^x}.$$
MyApproach:
Can this be solved L'Hospitals theorem? My problem is that L'Hospitals can be used when numerator and denominator is differentiated with the same variable but here $x$ and $y$ are both tending to $0$ so I don't think this theorem can be applied here. Is there any other way to solve it?
You are correct, L'Hopital's rule can be applied here. Actually, the limit does not even exist because by evaluating the limit along the sequences $(x_n,y_n)=(1/n,1/n)$ and $(x_n,y_n)=(1/n,e^{-n})$, which go to $(0^+,0^+)$ as $n$ goes to infinity, we obtain different results.
As regards $(x_n,y_n)=(1/n,1/n)$: \begin{align*} \frac{x_n^{y_n^{x_n}}-x_n^{y_n^{2}}}{y_n^{x_n}}&= \frac{\exp(\ln(x_n)\exp(x_n\ln(y_n)))-{\exp(y^2_n\ln(x_n))}}{\exp(x_n\ln(y_n))}\\ &\to\frac{\exp(-\infty\cdot e^0)-e^0}{e^0}=\frac{0-1}{1}=-1. \end{align*}
On the other hand for $(x_n,y_n)=(1/n,e^{-n})$: \begin{align*} \frac{x_n^{y_n^{x_n}}-x_n^{y_n^{2}}}{y_n^{x_n}}&= \frac{\exp(\ln(x_n)\exp(x_n\ln(y_n)))-{\exp(y^2_n\ln(x_n))}}{\exp(x_n\ln(y_n))}\\ &\to\frac{\exp(-\infty\cdot e^{-1})-e^0}{e^{-1}}=\frac{0-1}{e^{-1}}=-e. \end{align*}