How to write the definition of $\limsup_{(u, v)\to(0, 0)} \frac{f(x, u, v)}{\vert u\vert^{p} +\vert v\vert^q} < \lambda$? I am trying to write "explicity" what the expression
$$\limsup_{(u, v)\to(0, 0)} \frac{f(x, u, v)}{\vert u\vert^{p} +\vert v\vert^q} < \lambda$$
means. I proceed in this way:
$$\forall\varepsilon > 0 \ \ \exists \ \delta_{\varepsilon} >0 \mbox{ such that } \vert(u, v)\vert <\delta_{\varepsilon}: \quad f(x, u, v) < \lambda(\vert u\vert^p +\vert v\vert^q) -\varepsilon.$$
I would like to know if what I’ve written is true. I am a little bit doubtful because of the two variables.
Could anyone please help me?
Thank you in advance!
 A: *While I do not know what your $|(u,v)|$ is, I strongly believe that it is anyway a norm (probably a 2-norm of $\infty$-norm) and I will answer assuming that.
$$\limsup_{(u,v)→(0,0)}\frac{f(x,u,v)}{|u|^p+|v|^q}<λ\Leftrightarrow \lim_{r\rightarrow 0}\sup_{0<|(u,v)|<r}\frac{f}{|u|^p+|v|^q}<\lambda$$
$\Leftrightarrow\exists L<\lambda$ s.t. $\forall\epsilon>0,\exists\delta>0$ s.t. $\forall u,v\in\mathbb{R}$ s.t $0<|(u,v)|<\delta,\frac{f}{|u|^p+|v|^q}\le L+\epsilon$
A: Let $y=(u,v)$ and let $g(y)=\frac{f(x,u,v)}{|u|^p+|v|^q}$.
Consider the definition of $\limsup$:
$$\limsup_{y\to 0} g(y) = \lim_{r\to 0}\left\{\sup_{|y|\le r} g(y)\right\}$$
And consider the definition of $\lim$:
$$\lim_{r\to 0} g(r) = L \iff\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|r|<\delta_\epsilon: |g(r)-L|<\epsilon$$
Now let $L=\lim\limits_{r\to 0} g(r)$.
Working through your expression, we get:
$$\lim\limits_{r\to 0} g(r) = L <\lambda \\
\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|r|<\delta_\epsilon: |g(r)-L|<\epsilon\quad\land\quad L<\lambda \\
\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|r|<\delta_\epsilon: L-\epsilon < g(r) <L+\epsilon\quad\land\quad L<\lambda \\
\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|r|<\delta_\epsilon: L-\epsilon < \sup_{|y|\le r}g(y) <L+\epsilon\quad\land\quad L<\lambda
$$
If we assume that the limit exists, this simplifies to:
$$\text{limit exists}\quad\land\quad\exists L<\lambda\,\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|r|<\delta_\epsilon: g(r) \le L+\epsilon$$
Writing it with your original expression gives:
$$\text{limit exists}\quad\land\quad\\\exists L<\lambda\,\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|(u,v)|<\delta_\epsilon: \frac{f(x,u,v)}{|u|^p+|v|^q} \le L+\epsilon \\
\text{limit exists}\quad\land\quad\\\exists L<\lambda\,\forall \epsilon>0 \,\exists \delta_\epsilon>0 \text{ such that } 0<|(u,v)|<\delta_\epsilon: {f(x,u,v)} \le (L+\epsilon)({|u|^p+|v|^q})
$$
As you can see, this is different from what you wrote.
