Continuity of a function with real paramter Let $f:\mathbb{R^3}$ $\rightarrow$ $\mathbb{R}$,  defined as: 
$$f(x,y,z)=\begin{cases} \left(x^2+y^2+z^2\right)^p \exp\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)& ,\,\text{if }\quad(x,y,z) \ne (0,0,0)\quad \\
 0 &,\,\text{o.w}
\end{cases}$$
Where $\,p\in \mathbb{R}$. Is this function is continuous?

$f$ is continuous in $\mathbb{R}^3\setminus\{(0,0,0)\}$, but not in the point $(0,0,0)$ since the limit of $f(x,y,z)$ when $(x,y,z)\to(0,0,0)$ does not exist: To see it recall that
$$
e^{t}=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots,
$$
so for $t>0$ we have
$$
e^{t}\geq\frac{t^{2p+2}}{(2p+2)!}.
$$
Plugging $t=\sqrt{\frac{1}{x^2+y^2+z^2}}$ and translating in terms of $f$ this estimate gives
$$
f(x,y,z)\geq\frac{(x^2+y^2+z^2)^p}{(x^2+y^2+z^2)^{p+1}(2p+2)!}=\frac{1}{(x^2+y^2+z^2)(2p+2)!}\overset{(x,y,z)\to0}{\to}\infty.
$$
