# Proving that an Epsilon-Delta Proof is not true

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.
• Your negated formula is not correct: you should have $|f(x)-1/2|\geq\epsilon$ in the final bit. – Leo163 Nov 30 '18 at 12:20

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

Let $$\epsilon = \frac14$$, then $$\forall \delta >0$$, let $$x= \frac{\delta}2$$, then $$f(x)-\frac12=1-\frac12 \ge \epsilon$$
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 $$f(x_0)=0$$.