Proof that that if a function $f$ is continuous at some point, then multiplying $f$ by a constant does not affect its continuity. I want to show that if a function $f$ is continuous at some point, then multiplying $f$ by a constant does not affect its continuity.
Here's what I have:
Define the function $h (x) = nf(x),n\in\mathbb{R}$.
If $f$ is continuous at $x = a$, then, by the definition of continuity, there
is a cutoff distance $\delta > 0$ such that if $| x - a | < \delta$, then $| f
(x) - f (a) | < \epsilon$ for every positive tolerance $\epsilon > 0$.
For $h$ to be continuous at $x = a$, we must have that for every positive
tolerance $\epsilon_2 > 0$, there is a cutoff distance $\delta_2 > 0$ such
that if $| x - a | < \delta_2$, then
\begin{align}
  | h (x) - h (a) | < & \epsilon_2 \nonumber\\
  | n f (x) - n f (a) | < & \epsilon_2 \nonumber\\
  n | f (x) - f (a) | < & \epsilon_2 \nonumber\\
  | f (x) - f (a) | < & \frac{\epsilon_2}{n} . \nonumber
\end{align}
Since $\frac{\epsilon_2}{n} > 0$ and $f$ is continuous at $x = a$, the
inequality holds and $h$ is continuous at $x = a$.
 A: Your basic idea is correct, apart from the zero thing, but you need to add more details because as it stands, some sentences in your proof are untrue. First, in your big block of equations, you need to explain the relationship between the lines. The definition of continuity doesn't require all of the lines to be true, it only requires the first line to be true. You should make this clear, and then say that each of the following lines is equivalent to the first line (or say "By definition of $h$, it suffices to prove ...").
Also, it isn't true that "since $f$ is continuous, the inequality holds". The inequality doesn't hold unless you also assume $x < \delta_2$ for a suitable $\delta_2$, and you have to say where $\delta_2$ comes from. You should say, "Since $f$ is continuous, there exists some $\delta_2>0$ such that $|f(x)-f(a)|<\frac{\epsilon_2}{n}$ for all $x \in \mathbb{R}$ such that $|x-a|<\delta_2$."
Lastly, as a point of clarity, I would recommend not using $n$ (or $i$, $j$, etc.) for a real number - they are conventionally reserved for integers. It doesn't make your proof wrong, but it does make your proof less clear, and if you're being marked on clarity or mathematical communication, I would recommend changing it (to e.g. $c,\alpha,\beta,\eta,\lambda,\mu,a,b,\ldots$).
