# calculate limit of $\lim\limits_{(x,y)\to(0,0)} \frac{e^\frac{-1}{x^2+y^2}}{x^6+y^6}$

I need to calculate this limit: $$\lim\limits_{(x,y)\to(0,0)} \cfrac{1}{x^6 + y^6} exp \Biggr(\cfrac{-1}{x^2 + y^2} \Biggl)$$

It should be zero. I thought about the sandwich theorem but I'm not sure what to do from here:

$$0\leq \Biggr|\cfrac{1}{x^6 + y^6} exp \Biggr(\cfrac{-1}{x^2 + y^2} \Biggl) \Biggr|$$

## 1 Answer

Our goal is to show $$\lim\limits_{(x,y)\to(0,0)}\;\left(\frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}\right)\!=0$$ Changing to polar coordinates, let $$\begin{cases} x=r\cos(\theta)\\[4pt] y=r\cos(\theta)\\ \end{cases}$$ Noting that for $$\theta\in [0,2\pi]$$, the function $$\theta \mapsto \cos^6(\theta)+\sin^6(\theta)$$ has positive minimum value, $$a$$ say, we get $$0 < \frac {e^{-\frac{1}{x^2+y^2}}} {x^6+y^6} = \frac {e^{-\frac{1}{r^2}}} {r^6\bigl(\cos^6(\theta)+\sin^6(\theta)\bigr)} \le \frac {e^{-\frac{1}{r^2}}} {ar^6}$$ for all $$(x,y)\in\mathbb{R}^2{\setminus}\{(0,0)\}$$.

Hence, to prove the desired limit, it suffices to prove $$\lim_{r\to {0^{+}}} \frac {e^{-\frac{1}{r^2}}} {r^6} = 0$$ Then, letting $$u=\frac{1}{r^2}$$, we get $$\lim_{r\to {0^{+}}} \frac {e^{-\frac{1}{r^2}}} {r^6} = \lim_{u\to \infty} \frac{u^6}{e^u} = 0$$ as was to be shown.