Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$

Question

I'm looking for assistance with the following problem:

Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ where $f(x)=x^2-1$.

My attempt:

We need $\delta$ such that $\vert f(x)-2 \vert < .2$. That is, we need $\vert x^2 - 1 - 3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < .2$. We can control $\vert x - 2 \vert$ by $\delta$. Let us then restrict $x$ to the interval $(1,3)$. Thus $\vert x+2 \vert < 5$. Then we will let $\delta = \frac{.2}{5}$.

Now then, if $0 < \vert x - 2 \vert < \delta = \frac{.2}{5}$ then $\vert f(x)-3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < 5 \cdot \frac{.2}{5}=.2.$ $\Box$

Does this work for a valid delta? I'm basically taking the more general case where $\delta$ would be $\min(\frac{\epsilon}{5},1)$ and trying to fit it to this specific case.

Answer 1

Since \begin{eqnarray} |f(x)-3|&=&|x^2-4|\\ &=&|x-2||(x-2)+4|\\ &\le& |x-2|(|x-2|+4)\\ &=&|x-2|^2+4|x-2| \end{eqnarray} if we solve the inequality $$ y^2+4y\le 0.2 $$ we get $$ a\le y\le b, $$ with \begin{eqnarray} a&=&-2-\sqrt{4.2}\approx-4.0494\\ b&=&-2+\sqrt{4.2}\approx0.0494\\ \end{eqnarray} Hence, choosing $\delta \in (0,b)$, we have $$ 0<|x-2|<\delta \implies a<|x-2|<b \implies |f(x)-1|\le |x-2|^2+4|x-2|<0.2 $$

Answer 2

The other solution is of course neat :-)

If $|x^2-4|<0.2$ then we have that $3.8<x^2<4.2$ that is either

1. $\sqrt{3.8}<x<\sqrt{4.2}$ or

2. $-\sqrt{4.2}<x<-\sqrt{3.8}$.

Consider case $1$. From here we get \begin{align} -0.05064\sim\sqrt{3.8}-2&<x-2<\sqrt{4.2}-2\sim0.04939 \end{align} Now if we want $-\delta<x-2<\delta$, then we can choose $\delta<\sqrt{4.2}-2$.

For case $2$ we have \begin{align} -4.04939\sim-\sqrt{4.2}-2&<x-2<-\sqrt{3.8}-2\sim -3.94936 \end{align} implying that $\delta$ shall be negative which is not possible.