Prove that $(x^2+2x+1)$ is continuous by using a $\epsilon$ - $\delta$ argument

Question

I want to prove that the following function is continuous with an Epsilon Delta Argument. $$ f: \mathbb R \rightarrow \mathbb R \; (x^2+2x+1)$$

So I started with $$\left\lvert f(x)-f(y) \right\rvert= \left\lvert (x^2+2x+1) - (y^2+2y+1) \right\rvert = \left\lvert (x^2+2x)-(y^2+2y) \right\rvert$$

$$\le 2\left\lvert \left\lvert \frac{x^2-y^2}{2} \right\rvert + \left\lvert x- y\right\rvert \right\rvert $$ $$\le 2\left\lvert \left\lvert \frac{(x+y)(x-y)}{2} \right\rvert + \delta \right\rvert$$ $$= 2\left\lvert \frac{\left\lvert(x+y)(x-y)\right\rvert} {2} + \delta \right\rvert $$ $$ \le 2\left\lvert \frac{\left\lvert(x+y)\delta\right\rvert} {2} + \delta \right\rvert $$ $$= 2\left\lvert \frac{[\left\lvert(x-y)\right\rvert+\left\lvert2y\right\rvert]\delta} {2} + \delta \right\rvert $$ Now I define $\delta \le1.$ $$= 2\left\lvert \frac{(1+\left\lvert2y\right\rvert)\delta} {2} + \delta \right\rvert $$ $$= \left\lvert(1+\left\lvert2y\right\rvert)\delta + 2\delta \right\rvert $$ $$= \delta \left\lvert(1+\left\lvert2y\right\rvert) + 2 \right\rvert $$ $$= \delta \left\lvert\left\lvert2y\right\rvert + 3 \right\rvert = \epsilon $$ And so my $\delta$ is $\frac{\epsilon}{\left\lvert\left\lvert2y\right\rvert + 3 \right\rvert}$ and $\delta \le 1$

Can anyone tell my if this is correct or where I went wrong I am studying for exam. :) Thanks in advance.

Answer 1

I don't know why you want to factor the number $2$ out of your estimate. But the following may be a more natural way to think about how to find a needed $\delta$.

Notice that $$ |x^2+2x-y^2-2y|\le|x+y||x-y|+2|x-y|=(|x+y|+2)|x-y| \tag{1} $$

Suppose you want to show that your function is continuous at $y$. If $|x-y|<\delta$, you want $$ (|x+y|+2)|x-y|\le \epsilon. \tag{2} $$

On the other hand,

$$ (|x+y|+2)|x-y|\le (|x-y|+2|y|+2)|x-y|\le (\delta+2|y|+2)\delta\tag{3} $$

So if you let $\delta<1$ and $(2|y|+3)\delta<\epsilon$, you get what you want.

To wrap things up, set $\delta=\min(1,\frac{\epsilon}{2|y|+3})$.

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Your solution is ok except for several places. See below.

So I started with (*assuming $|x-y|<\delta$ *) $$\left\lvert f(x)-f(y) \right\rvert= \left\lvert (x^2+2x+1) - (y^2+2y+1) \right\rvert = \left\lvert (x^2+2x)-(y^2+2y) \right\rvert$$

$$\le 2\left\lvert \left\lvert \frac{x^2-y^2}{2} \right\rvert + \left\lvert x- y\right\rvert \right\rvert $$ $$< 2\left\lvert \left\lvert \frac{(x+y)(x-y)}{2} \right\rvert + \delta \right\rvert$$ $$= 2\left\lvert \frac{\left\lvert(x+y)(x-y)\right\rvert} {2} + \delta \right\rvert $$ $$\color{red}{\le} 2\left\lvert \frac{\left\lvert(x+y)\delta\right\rvert} {2} + \delta \right\rvert $$ $$= 2\left\lvert \frac{[\left\lvert(x-y)\right\rvert+\left\lvert2y\right\rvert]\delta} {2} + \delta \right\rvert $$ Now if I set $\delta \le1$ then $$= 2\left\lvert \frac{(1+\left\lvert2y\right\rvert)\delta} {2} + \delta \right\rvert $$ $$= \left\lvert(1+\left\lvert2y\right\rvert)\delta + 2\delta \right\rvert $$ $$= \delta \left\lvert(1+\left\lvert2y\right\rvert) + 2 \right\rvert $$ $$= \delta \left\lvert\left\lvert2y\right\rvert + 3 \right\rvert = \epsilon $$ And so my $\delta$ is $\min(1,\frac{\epsilon}{\left\lvert2y\right\rvert + 3 })$.

Answer 2

As a comparison I did this

$$ \left| x^2 + 2x + 1 - y^2 - 2y - 1 \right| < |x-y|^2 + (2|y| + 2)|x - y| < \epsilon $$

And I defined $\delta$ as the strictly positive solution of $\delta^2 + (2|y| + 1)\delta = \epsilon$, which you can easily show it exists for every $\epsilon > 0$.