How do I get rid of the $x$ in the (Epsilon-Delta-Definition)

Question

$f(x)= \left\{ \begin{align} 2x+1 \quad \text{if } x\in \mathbb{Q} \\ -2x^2+3x+4 \quad \text{if } x \notin \mathbb{Q} \end{align} \right.$

Show that $f$ is continuous in $x_0=-1$

Here's my attempt:

(i) $x\in \mathbb{Q}$

For every $\varepsilon >0$, find $\delta_1>0$, such that for all $x_0, x\in \mathbb{Q}$: $|x-x_0|<\delta_1 \implies |f(x)-f(x_0)|<\varepsilon $

$\Leftrightarrow |x+1|<\delta_1 \implies |2x+2|<\varepsilon$

Since $|2x+2|=2|x+1|<\varepsilon$, choose $\delta_1 := \varepsilon /2$.

(ii) $x\notin \mathbb{Q}$

For every $\varepsilon >0$, find $\delta_2 >0$, such that for all $x,x_1\notin \mathbb{Q}$ gilt:

$|x+1|<\delta \implies |-2x^2+3x+5|<\varepsilon $

We have $|-2x^2+3x+5|=|-(x+1)\cdot (2x-5)|=|x+1|\cdot |2x-5|<\delta \cdot |2x-5|$.

How do I get rid of the $x$?

Answer 1

Hint

when $|x+1|<\delta$ then for sufficiently small $\delta>0$ we have $$6<|2x-5|<8$$

Answer 2

In general if you get something in terms of $|something| < c*\delta + k*\delta^2$ you can presume that if $\delta < 1$ then $\delta^2 < \delta$ and you have $c*\delta + k*\delta^2 \le c*\delta + k*\delta = \delta(c+k)$.

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If $|x + 1| < \delta$ and $x$ is rational then $|f(x) - f(-1)| = |2x + 1 - (-1)| = |2x+2| = 2|x+1|< 2\delta$.

If $x$ is irrational then $|f(x) -f(1)| =|-2x^2 + 3x +4 -(-1)|=|-2x^2 +3x +5| = |x+1||5-2x| < |5-2x|\delta$.

Now $|x+1| < \delta$ means

$-\delta < x+1 < \delta$

$-1 -\delta < x < -1+\delta$

$2-2\delta < -2x< 2+2\delta$ and

$7-2\delta < 5-2x < 7+2\delta$. If we presume $\delta < 3.5$ we have

$0 < 5-2x = |5-2x| < 7+2\delta$ and so

$|f(x) - 1| = |x+1||5-2x|<|5-2x|\delta < (7+2\delta)*\delta =7\delta + 2\delta^2$.

If we presume $\delta < 1$ then we can either reason $7+2\delta < 9$ or that $\delta^2 < \delta$ so $2\delta^2 < 2\delta$. Either way.

$|f(x) -f(1)|< 9\delta$.

So for any $\epsilon > 0$ if we set $\delta = \min(1, \frac {\epsilon}9)$ then

If $|x-(-1)|< \delta\le \frac {\epsilon}9\$ and $x$ is rational then

$|f(x) -f(-1)|= 2|x+1| < 2\delta \le \frac {2\epsilon}9 < \epsilon$.

If $x$ is irrational then

$|f(x) - f(-1)|=|x+1||5-2x|<|x+1|*9 \le 9\frac {\epsilon}9=\epsilon$.

In general, restricting $\delta < 1$ was not as tight as we could do. We could restrict $\delta < \frac 1n$ let $\delta =\min(\frac 1n,\frac {\epsilon}{7+\frac 2n})$ will do for any $n$ but... why go through the trouble.