Proving that an Epsilon-Delta Proof is not true

Question

We write $\lim_{x \to a} f(x)=L$ if the following is true $(\forall\epsilon>0)(\exists\delta>0)(\forall x)(0<|x-a|<\delta\rightarrow|f(x)-L|<\epsilon)$

Let $f:\Bbb{R}\rightarrow\Bbb{R}$ be given by

$f(x)= \begin{cases} 0, & \text{if $x<0$} \\ 1/2, & \text{if $=0$} \\ 1, & \text{if $x>0$} \end{cases}$

We will show that it is not the case that $\lim_{x \to 0} f(x)=1/2$

(a) Write the negation of $\lim_{x \to 0} f(x)=1/2$ using the epsilon-delta definition given above

• I attempted to find the negation of this and this what I got after some calculations

$(\exists\epsilon>0)(\forall\delta>0)(\exists x)(0<|x-0|<\delta\land|f(x)-1/2|\ge\epsilon)$

(b) Prove the assertion that you found in part (a) Hint: $\epsilon=1/4$

• This is where I am stuck. How would I prove the epsilon-delta expression that I found in part (a). Any kind of help would be appreciated.

Answer 1

$$(\exists\epsilon>0)(\forall\delta>0)(\exists x)(0<|x-0|<\delta\land|f(x)-1/2|\color{blue}\ge\epsilon)$$

Follow the hint,

Let $\epsilon = \frac14$, then $\forall \delta >0$, let $x= \frac{\delta}2$, then $$f(x)-\frac12=1-\frac12 \ge \epsilon$$

Answer 2

Given $\delta>0$, we look for $x$ such that

$$0<|x|<\delta $$ and $$|f(x)-\frac 12|\ge \frac 14.$$

$$\iff$$

$$f(x)\ge \frac 34 \text{ or } f(x)\le \frac 14$$

so we can take $x_0=-\frac{\delta}{2}$ with $f(x_0)=0$.