Show that $f(x)=x^2+2$ is continuous at the point $x=1$.
Can you provide feedback on my proof of the above claim?
Proof. Let $\epsilon>0$, and choose $\delta=\frac{\epsilon}{|x+1|}$ then for all $x\in\mathbb{R}$ we have,
$|x-1|<\delta=\frac{\epsilon}{|x+1|}\Rightarrow |x-1||x+1|<\epsilon \Rightarrow |x^2-1|=|x^2+2-1|<\epsilon$.
Hence, for every $\epsilon>0$ there is a $\delta>0$ such that for all $x\in\mathbb{R}$ with $|x-1|<\delta$, we have $|f(x)-f(1)|<\epsilon$.
Q.E.D.
Method 1 (the way you should learn): I'm going to change the numbers around so you have to work through your problem. Show $f(x)=x^2-3$ is continuous at $x=2$.
Scratch work: It looks like $f(2)=1$. Then $|f(x)-f(1)|=|x^2-4|=|x+2|\,|x-2|$. Obviously, the $|x+2|$ is the weird term. Say it wasn't there and it was just constant $C>0$. Then for $\epsilon>0$, choose $\delta=\frac{\epsilon}{C}$. Then $C|x-2|<C\left(\frac{\epsilon}{C}\right)=\epsilon$. This is why you (incorrectly) wanted to make $C=\frac{\epsilon}{|x+2|}$ because you also wanted to get rid of it. Unfortunately, $\delta$ has to be fixed and independent of $x$.
Now we know we want to get rid of $|x+2|$. Well in the $\epsilon$-$\delta$ definition, once we get a $\delta$, we an always restrict $\delta$ by making it smaller.
To start, choose $\delta_1=1$. In this neighborhood, $|x-2|<\delta_1 \implies -\delta_1<x-2<\delta_1\implies -\delta_1+4<x+2<\delta_1+4$ so $3<x+2<5$. Hence, we can say $|x+2|<5$.
Now the problem is familiar to us. Choose $\delta_2=\frac{\epsilon}{5}$ and $\delta=\min\{\delta_1,\,\delta_2\}$. Then when $|x-2|<\delta$, we have $|x-2|<\delta_1$ so $|x+2|<5$. Moreover,$|x-2|<\delta_2$. Hence, when $x$ satisfies $|x-2|<\delta$, we have $|f(x)-f(1)|=|x+2|\,|x-2|<5\left(\frac{\epsilon}{5}\right)=\epsilon$.
Method 2. There's a theorem that says:
The function $f(x)$ is continuous at $x=a$ if and only if for every sequence $x_n$ with $x_n \to a$, it follows that $f(x_n) \to f(a)$.
Then the problem can be treated like the standard $\epsilon-N$ problems for sequences.