Proving continuity in multivariable calculus using $\epsilon - \delta$ definition The textbook im using only has stated the definition of limit with examples explained intuitively and without full rigour, so im pretty lost on multivariable limit/continuity problems. I have the following problem at hand and I have to prove that its continuous using $\epsilon - \delta$ definition: given $a_1, \ldots, a_n \in \mathbb{R^n}$ and a map $g: \mathbb{R^n} \rightarrow \mathbb{R}$ defined as $$g(a_1, \ldots, a_n) = a_1x_1 + a_2x_2 + \cdots + a_nx_n$$ Now the right side of the equation looks like a polynomial and I know that they are always continuous everywhere. But I cannot use any other theorems to prove it since it wouldnt be $\epsilon - \delta$. Some theorem I couldve used to make it easier is by taking coordinate function wise limit, but that does not help with the goal.
Essentially I have to show that :
$\forall \epsilon > 0, \exists \delta > 0$ such that $x \in \mathbb{R^n}$ and $0 < |x - a| \implies  |g(a_1, \ldots, a_n) - (a_1x_1 + a_2x_2 + \cdots + a_nx_n)| < \epsilon$
Would appreciate any insight or general approach to solving multivariable limit/continuity problems
 A: We want to show that $g$ is continuous everywhere, so pick some arbitrary point $a = (a_1, ..., a_n)$.
Now consider some arbitrary $\varepsilon > 0$ and our goal becomes showing that there exists some $\delta > 0$ such that for all $a' = (a'_1, ..., a'_n)$ with $\|a - a'\| < \delta$ we have $|g(a) - g(a')| < \varepsilon$.
Expanding this, what we want is that
$$|g(a) - g(a')| = |a_1 x_1 + ... + a_n x_n - (a'_1 x_1 + ... + a'_n x_n)| = |(a_1 - a'_1)x_1 + ... + (a_n - a'_n) x_n| < \varepsilon.$$
Define $X = |x_1| + ... + |x_n|$ and let $\delta = \varepsilon / X$.
We know that the euclidean norm is bounded by the sum of absolute differences, i.e. if we have some $a'$ such that $\|g(a) - g(a')\| < \delta$ then we can conclude
$\|g(a) - g(a')\| \leq |a_1 - a'_1| + ... + |a_n - a'_n|$, so we get that $|a_i - a'_i| < \delta$ for all $i \in [1, n]$.
Using this and the triangle inequality, we find
$$
\begin{align*}
|g(a) - g(a')|
&= |(a_1 - a'_1)x_1 + ... + (a_n - a'_n) x_n| \\
&\leq |(a_1 - a'_1)| \cdot |x_1| + ... + |(a_n - a'_n)| \cdot |x_n| \\
&< \delta |x_1| + ... + \delta |x_n| \\
&= \delta \cdot (|x_1| + ... + |x_n|) \\
&= \delta \cdot X \\
&= \varepsilon
\end{align*}
$$
and thus it follows that $g$ is continuous at $a$.
As $a$ was arbitrary, $g$ is continuous everywhere.
A: I am assuming that the variable is a vector $x=(x_1,x_2...x_n)$ and you have to show continuity near $x=x_0=(x_{01},x_{02}....x_{0n})$. To show that we need to show that we can make $|g(x)-g(x_0)|$ as small as we want by making $|x-x_0|$ as small as we need to. Thus given $\epsilon$ we want to make the following to be small than $\epsilon$:
$$|g(x)-g(x_0)|=|\sum_{i=1}^n a_ixi-\sum_{i=1}^n a_ix_{0i}|=\sum_{i=1}^na_i(x_i-x_{i0}|\le \sum_{i=1}^n|a_i||x_i-x_{i0}|
$$
To achieve it we choose $\delta\lt\frac{\epsilon}{n \max{|a_i|}}$ and we get:
$$
|g(x)-g(x_0)|\lt\sum_{i=1}^n|a_i|\frac{\epsilon}{n \max{|a_i|}}=\frac{\epsilon}{n}\sum_{i=1}^n\frac{|a_i|}{ \max{|a_i|}}\le\epsilon
$$
A: Often for multivariable problems like this (needing to show continuity of $f$ at a point $(b_1,b_2,\dots,b_n)$), given an $\epsilon$, one can first find a $\delta_i$ for each coordinate, $i$, such that whenever $|x_i-b_i|<\delta_i$, then $|f(b_1,b_2,\dots, x_i,b_{i+1},\dots b_n)-f(b_1,b_2,\dots, 
b_i,b_{i+1},\dots b_n)|<\textrm{some expression involving}\; \epsilon$. Then, it often turns out that any $\delta$ small enough compared to all the $\delta_i$s (e.g. $\leq \textrm{min}(\delta_1,\delta_2,\dots,\delta_n)$) does the job for the entire function, i.e. for that suitably small $\delta$ we in fact have that $|(x_1,x_2,\dots,x_n)-(b_1,b_2,\dots,b_n)|<\delta$ implies $|f(x_1,x_2,\dots,x_n)-f(b_1,b_2,\dots,b_n)|<\epsilon$.
Regarding the question at hand - suppose you are given some $\epsilon>0$ and some point $(b_1,b_2,\dots,b_n)$. The way you can apply the idea in the above paragraph here, is to first, using the fact that each single variable function $g(b_1,b_2,...,x_i,b_{i+1},\dots,b_n)=a_1b_1+a_2b_1+\dots+a_ix_i+a_{i+1}b_{i+1}+\dots+a_nb_n$ is continuous, we can pick a $\delta_i$ such that $|g(b_1,b_2,...,x_i,b_{i+1},\dots,b_n)-g(b_1,b_2,...,b_i,b_{i+1},\dots,b_n)|<\frac{\epsilon}{n}$, i.e. $|x_i-b_i|<\delta_i$ implies that $|a_ix_i-a_ib_i|<\frac{\epsilon}{n}$. It is now very easy to show via the triangle inequality that, setting $\delta=\textrm{min}(\delta_1,\delta_2,\dots,\delta_n)$, whenever $|(x_1,x_2,\dots,x_n)-(b_1,b_2,\dots,b_n)|<\delta$, then $|g(x_1,x_2,\dots,x_n)-g(b_1,b_2,\dots,b_n)|<\epsilon$, completing the epsilon-delta proof of continuity.
