Epsilon delta proof for the limit of x squared plus 1 at x equals 1

Key takeaway: The heart of an $\epsilon$-$\delta$ proof is always the same: turn the expression $|f(x)-L|$ into something controlled by $|x-a|$. For this function, the algebra is friendly, so the proof is short once the right estimate is chosen.

We want to prove that

$\lim_{x\to 1}(x^2+1)=2.$

Let $\epsilon>0$ be given. We must find $\delta>0$ such that

$0<|x-1|<\delta \implies |(x^2+1)-2|<\epsilon.$

Start by simplifying:

$|(x^2+1)-2|=|x^2-1|=|x-1||x+1|.$

Now restrict $x$ to be near 1. If $|x-1|<1$, then $0<x<2$, so

$|x+1|<3.$

Thus when $|x-1|<1$,

$|(x^2+1)-2|=|x-1||x+1|<3|x-1|.$

To make this less than $\epsilon$, it is enough to require

$3|x-1|<\epsilon.$

So choose

$\delta=\min\left(1,\frac{\epsilon}{3}\right).$

Then if $0<|x-1|<\delta$,

$|(x^2+1)-2|<3|x-1|<3\cdot\frac{\epsilon}{3}=\epsilon.$

Therefore,

$\lim_{x\to 1}(x^2+1)=2.$

---

Pitfalls the pros know 👇 A typical error is to set $\delta=\epsilon$ without bounding the extra factor $|x+1|$. In an $\epsilon$-$\delta$ proof, that factor matters. You need a small neighborhood around the point first, then a second bound that translates the expression into something directly controlled by $|x-1|$.

What if the problem changes? If the limit point were $x\to a$ instead of $x\to 1$, the same pattern works:

$|x^2+1-(a^2+1)|=|x-a||x+a|.$

You then bound $|x+a|$ by forcing $x$ close enough to $a$, and choose $\delta$ as the minimum of that neighborhood size and the amount needed to make the final estimate smaller than $\epsilon$.

Tags: epsilon delta, limit definition, continuity