Alternative multivariable limit definition According to the definition of a limit in a multivariable function we have:

Suppose $f:U\to R$, where $U$ open set, $U\subseteq R^n$ and $\boldsymbol{x_0}\in U$. We say that $lim_{\boldsymbol{x}\to \boldsymbol{x_0}}f(\boldsymbol{x}) = a$ if $\forallε>0$, $\exists δ>0$ such that $$0<||\boldsymbol{x} - \boldsymbol{x_0}|| < δ \implies |f(\boldsymbol{x}) - a| < ε$$

Let's take the same function $f:U\to R$ where $U\subseteq R^n$. Therefore the function has the following form
$$f(\boldsymbol{x}) = f(x_1, x_2, \dots, x_n)$$
I came up with the following definition of the multivariable limit:

Suppose $f:U\to R$ where $U$ open set, $U\subseteq R^n$ and $\boldsymbol{x_0} = (x_{01}, x_{02}, \dots, x_{0n})\in U$. We say that $lim_{\boldsymbol{x}\to \boldsymbol{x_0}}f(\boldsymbol{x}) = a$ if $\forallε>0$, then $\exists δ_1>0, \exists δ_1>0, \dots \exists δ_n>0$ such that if $$0<|x_1 - x_{01}| < δ_1$$ $$0<|x_2 - x_{02}| < δ_2$$ $$\dots$$ $$0<|x_n - x_{0n}| < δ_n$$ then $$|f(\boldsymbol{x}) - a| < ε$$

Is that a correct definition? Is that definition equivalent with the one introduced in any calculus book?
 A: It's best to view this in terms of equivalence of norms. If you have a norm $\|\cdot\|$ as you do in your question, $\|{\bf x-y}\|$ measures a distance between the points $\bf x$ and $\bf y$. For $p\geq 1$, let $$\|(a_1,\cdots,a_d)\|_{p} := \left(\sum_{i=1}^n |a_i|^p\right)^{1/p}.$$
Also, for $p=\infty$, we define
$$\|(a_1,\cdots,a_d)\|_{\infty} := \max_{i=1,\dots,d}\{|a_i|\}.$$
Then $\|\cdot\|_{p}$ is a norm in $\mathbb{R}^d$, for $1\leq p\leq \infty$. Therefore, you can think of $\|{\bf x-y}\|_{p} = \left(\sum_{i=1}^n |x_i-y_i|^p\right)^{1/p}$ as a $p$-distance of the two points $\bf x$ and $\bf y$. The $\|\cdot\|$ in the "known" definition is just $\|\cdot\|_2$. On the other hand, in the definition that you came up with, you are basically using $\|\cdot\|_\infty$ (take a second to see why is that. Hint: consider $\delta=\max\{\delta_1,\dots,\delta_n\}$).
We say that two norms are equivalent if there are constants $c,C$ such that $$ c \| {\bf x}\|_{p_1} \leq \|{ \bf x}\|_{p_2} \leq C \| {\bf x}\|_{p_1}$$
for every vector $\bf x$. The constants $c,C$ need to be universal, ie the same for all $\bf x$.
You can define the limit with respect to any norm: The definition is exactly the one you gave, just substitute $\|\cdot\|_{p}$ in the place of $\|\cdot\|$. If two norms are equivalent, then the definition of the limit is the same for the two norms. Intuitively, if $\|{ \bf x}\|_{p_2}$ is small, then so is $\| {\bf x}\|_{p_1}$, using the first inequality above, and vice versa using the second. It is a great exercise in the epsilon-delta definition to prove this formally (hint: Start with the definition for the one and use $\tilde{\delta}=\delta/c$ for the other).
Finally, it is an important fact that all the $p$-norms we defined above, are indeed equivalent! For any $1\leq p < q < \infty$,
$$  \| {\bf x}\|_{q} \leq \|{ \bf x}\|_{p} \leq n^{\frac{1}{p} - \frac{1}{q}} \| {\bf x}\|_{q}.$$
Note that the two constants (1 and $n^{\frac{1}{p} - \frac{1}{q}}$) are independent of $\bf x$. In particular:
$$  \| {\bf x}\|_{\infty}\leq  \| {\bf x}\|_{2} \leq \|{ \bf x}\|_{1}\leq \sqrt{n} \| {\bf x}\|_{2} \leq n\| {\bf x}\|_{\infty}.$$
Proving these inequalities yourself will help you understand why the equivalence holds.
