Checking proof that $f(x)=x^2+1$ is continuous Let $f:\mathbb R \to \mathbb R$ is defined by $f(x)=x^2+1$.  Prove this function is continuous for all $x \in\mathbb R$.
Here is what I have:
Suppose that $c∈ℝ$.  Let $\varepsilon>0$.  Let $\varepsilon=\delta$.  Then
$$
\begin{align}
& |f(x)-f(c)|=|x^2+1-(c^2+1)| \\[6pt]
= {} & |x^2-c^2| \\[6pt]
= {} & |x-c|^2 \\[6pt]
= {} & |x-c|< \sqrt{\varepsilon}
\end{align}
$$
Then 
$$|f(x)-f(c)|<(\sqrt{\varepsilon})^2=\varepsilon$$
Thus, proved.
 A: We want to show that for every $\epsilon\gt 0$ there is a $\delta\gt 0$ such that if $0\lt |x-c|\lt \delta$ then $|f(x)-f(c)|\lt \epsilon$.  Let $\epsilon\gt 0$ be given. We produce a suitable $\delta$.
We have 
$$|f(x)-f(c)|=|x^2-c^2|=|x+c||x-c|.\tag{1}$$
We want to show that the expression on the right side of (1) can be made "small" if we take $|x-c|$ sufficiently small.  
The potential "spoiler" is the term $|x+c|$, which could be large. There are two cases to consider, $c\ne 0$ and $c=0$. We leave the case $c=0$ to you , and deal with $c\ne 0$.  
We want to control the size of $|x+c|$. To do this, first specify that we will pick  $\delta \le |c|/2$. If $|x-c|\le |c|/2$, then $|x+c|\le 3|c|/2$. It follows that the expression on the right of (1) is $\le \frac{3}{2}|c||x-c|$.
Thus if $\delta$ satisfies the two conditions  $0\lt \delta\le |c|/2$ and $0\lt \delta \le \frac{2}{3|c|}\epsilon$, and if $|x-c|\lt \delta$, then we will have $|f(x)-f(c)|\lt \epsilon$. It is therefore enough to let $\delta=\min(|c|/2, \frac{2}{3|c|}\epsilon)$.
For the case $c=0$ that is left to you, we need to find a $\delta$ such that if $|x|\lt \delta$ then $x^2\lt \epsilon$. 
