We have the function $f:\mathbb{R}\to\mathbb{R}$, defined by $f(x)=x^2+3x-5$ and want to show that it is continuous at $3$. This is how I've done it
Proof:
Let $\varepsilon>0$. Take $\delta=\mbox{min}\{1,\frac{\varepsilon}{10}\}$ and suppose that $|x-3|<\delta$. We note that for $|x-3|<1$, we have $|x+6|=|x-3+9|\leq|x-3|+9<1+9=10$. Therefore, we have the following:
\begin{align*} |f(x)-f(3)|&=|x^2+3x-18|\\ &=|x-3||x+6|\\ &<\delta\cdot 10\\ &=\frac{\varepsilon}{10}\cdot10\\ &=\varepsilon\\ \end{align*} $\blacksquare$
I'm wondering if this proof is sound. If somebody could verify this for me I would greatly appreciate it. I'm still quite new to epsilon-delta proofs. Any additional advice or pointers are always appreciated as well. Thanks
From your second to last line... $\delta \cdot 10 \neq \frac{\varepsilon}{10} \cdot 10$. $$ \delta \cdot 10 = \begin{cases} \frac{\varepsilon}{10} \cdot 10 ,& \varepsilon/10 \leq 1 \\ 1 \cdot 10 ,& \varepsilon/10 > 1 \end{cases} \text{.} $$ The first case is, of course, $\varepsilon$, but the second is not.
In the second case, from $\varepsilon/10 > 1$, you have $\varepsilon > 10$, so $\delta \cdot 10 = 1 \cdot 10 < \varepsilon$ in this case.
If your audience would automatically understand this detail, you could correctly write $$ \delta \cdot 10 \leq \min\{1, \varepsilon/10\} \cdot 10 = \min\{10,\varepsilon\} \leq \varepsilon \text{.} $$