## Thomas' Calculus 13th Edition

Step 1. We can identify the quantities as: $c=0,L=0$. $f_1(x)=2x$ when $x\lt0$, and $f_2(x)=x/2$ when $x\geq0$. Step 2. For a given small value $\epsilon\gt0$, we need to find a value $\delta\gt0$ so that for all $x$ when $0\lt |x|\lt\delta$, we have $|f(x)|\lt\epsilon$ which gives either $|2x|\lt\epsilon$ ($x\lt0$) or $|x/2|\lt\epsilon$ ($x\geq0$) Step 3. The last two inequality gives $0\lt -2x \lt\epsilon$ or $-\frac{\epsilon}{2}\lt x \lt 0$. Also, $0\leq x/2 \lt\epsilon$ or $0\leq x \lt 2\epsilon$. These inequalities can be combined as $-\frac{\epsilon}{2}\lt x \lt 2\epsilon$. Step 4. From the requirement for the value $\delta$ in step 2, we have $-\delta\lt x\lt\delta$ Step 5. Compare the results from steps 4 and 3 ,we can set $\delta=\frac{1}{2}\epsilon$ Step 6. With the setting of the $\delta$ value, we can go backwards through the above steps to reach a conclusion that for all $x$ when $0\lt |x|\lt\delta$, we have $|f(x)|\lt\epsilon$ which proves the limit statement.