Continuity of multidimensional function $f(x,y) = \frac{x\exp\left({-1}/{y^2}\right)}{x^2+\exp\left({-2}/{y^2}\right)}$

Question

I have given the following task: Test if

$$f(x,y) = \begin{cases}\dfrac{x\exp\left(\frac{-1}{y^2}\right)}{x^2+\exp\left(\frac{-2}{y^2}\right)}&y\neq 0\\ 0&y=0\end{cases}$$ is continuous in $(0,0)$ using the parametric curve $\gamma(t) = (t^m,t^n)$ for $m,n \in \mathbb{N}$. That means I have to show that $\lim_{t\rightarrow 0}f(t^m,t^n) = (0,0)$.

So what I did is to calculate the limit $\dfrac{t^m\exp\left(\frac{-1}{t^{2n}}\right)}{t^{2m}+\exp\left(\frac{-2}{t^{2n}}\right)}$ for $t \rightarrow 0$ using L'Hospital's theorem. My question is if there is any trap in this task. Is there something I have to be aware of?

Can you give me a hint how to solve this more efficient? Please have in mind that I just learned about multidimensional continuity and dont know about the derivative of a multidimensional function at all.

Answer 1

Note that

$$\frac{t^me^{\frac{-1}{t^{2n}}}}{t^{2m}+e^{\frac{-2}{t^{2n}}}} =\frac{t^m}{t^{2m}e^{\frac{1}{t^{2n}}}+e^{\frac{-1}{t^{2n}}}}\to 0$$

indeed

$$t^{2m}e^{\frac{1}{t^{2n}}}=\frac{e^{\frac{1}{t^{2n}}}}{\frac1 {t^{2m}}}\to \infty$$

To show that $f(x,y)$ is not continuos at $(x,y)=(0,0)$ note that for $x=0$

$$\frac{xe^{\frac{-1}{y^2}}}{x^2+e^{\frac{-2}{y^2}}}=0$$

but for $x=t$ and $\left(-\frac1{y^2}\right)=\log t$ as $t\to 0^+$

$$\frac{xe^{\frac{-1}{y^2}}}{x^2+e^{\frac{-2}{y^2}}}=\frac{t^2}{2t^2}=\frac12$$

Answer 2

You're doing fine on this. I would just note that

$$|f(x,y)| \le \frac{|x|e^{-1/y^2}}{x^2 },$$

which will make your L'Hopital computations easier after letting $(x,y)=(t^m,t^n).$