4. Case 2 - No Limit

Find the $\delta$ & $\epsilon$ that satisfies

\lim_{x \to 0} \; f(x) = \begin{cases} 1 + x, & \text{if } x \ge 0,\\ -1 + x, & \text{if } x < 0 \end{cases}

As $x$ approaches $0$ from above $f(x)$ tends to $+1$

As $x$ approaches $0$ from below $f(x)$ tends to $-1$

If we pick a random value for $\epsilon$ say $\epsilon = 10^{-4}$ then

|f(x) - L| < 10^{-4}

Case 1: $x \ge 0$

|(1 + x) - L| < 10^{-4}

Here $L$ would have to be the limit as $x \to 0$ . For $x \ge 0$ , the function tends to 1, so

|1 + x - 1| < 10^{-4} \quad \to \quad 0 \le x < 10^{-4}

this approach works

Case 2: $x < 0$

|(-1 + x) - L| < 10^{-4}

if we try to pick $L = 1$ (the right hand limit), then

|-2 + x| \gg 10^{-4}

therefore no mater how small we pick $\delta$ , we cannot satisy the condition for $x < 0$