Prove that $f(x) = x^ 2 + 1$ is continuous on R using a δ − ε proof.

Question

For this problem, we weren't allowed to simply apply properties of continuous functions. My professor gave me some hints and tips but I'm still not sure how to finish out the proof. Here's what I have so far.

For ε, $|(x^2+1)-(a^2+1)|<ε$. He told me I want $|x-a|*|x+a|<ε$.

Then he said I should suppose δ=1 and $|x|=\leq 1+|a|$.

Next we did some work to end up with $|x-a|<\frac{ε}{2|a|+1}$.

And lastly, $δ=min(1,\frac{ε}{2|a|+1})$.

I've really been struggling with the δ and ε proofs and am not sure what to do next. Could someone work it through the rest of the way while explaining what they're doing? Thanks.

Answer 1

Please see my explanation of approaching these types of problems here. I will approach this using a similar method.

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Scratch Work

Let $\epsilon > 0$. We want to show $$|x^2 + 1 - (a^2 + 1)| = |x^2-a^2|=|x+a||x-a|<\epsilon\text{.}$$ We have control over $|x - a|$; suppose $|x - a| < 1$. Then $$|x - a| < 1 \Longleftrightarrow -1 < x - a < 1 \Longleftrightarrow a-1 < x < a + 1 \Longleftrightarrow 2a -1<x+a<2a+1\text{.} $$ This implies that $|x + a| < \max\{|2a-1|, |2a+1|\}=: M$. Furthermore, this implies that $$|x+a||x-a| < M|x-a|$$ so an appropriate $\delta$ is $\delta := \min\left(1, \dfrac{\epsilon}{M}\right)$.

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Proof

Let $\epsilon > 0$ and $a \in \mathbb{R}$ be arbitrary. Choose $\delta := \min\left(1, \dfrac{\epsilon}{M}\right)$, where $M:= \max\{|2a-1|, |2a+1|\}$. Then $$|x^2 + 1 - (a^2 + 1)| = |x^2-a^2|=|x+a||x-a| < M|x-a|\text{.}$$ This is because $$|x - a| < 1 \Longleftrightarrow -1 < x - a < 1 \Longleftrightarrow a-1 < x < a + 1 \Longleftrightarrow 2a -1<x+a<2a+1\text{.} $$ which implies that $|x + a| < \max\{|2a-1|, |2a+1|\}=: M$. Thus, $$|x^2 + 1 - (a^2 + 1)| \leq M|x-a| < M\delta \leq M\left(\dfrac{\epsilon}{M}\right) = \epsilon\text{.}$$

Answer 2

Fix $\varepsilon >0$ sufficiently small and $x_0 \neq 0$. It is enough to see that, if $|x-x_0|<\frac{\varepsilon}{\max(2,4|x_0|)}$ then $$ |f(x)-f(x_0)|=|x^2-x_0^2|<\frac{\varepsilon^2}{\max(4,16x_0^2)}+2|x_0|\frac{\varepsilon}{\max(2,4|x_0|)} $$ so that $$ |f(x)-f(x_0)|< \frac{\varepsilon^2}{4}+\frac{\varepsilon}{2}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. $$ On the other hand, for $x_0=0$, it is enough to choose $|x-x_0|<\sqrt{\varepsilon}$.