## Thomas' Calculus 13th Edition

(a) Step 1. Identify the given piece-wise function: $f(x)=\begin{cases}x,\hspace1.7cm x\lt1 \\ x+1,\hspace1cm x\gt1 \end{cases}$ Step 2. For each $\delta\gt0$, with $|x-1|\lt\delta$, we have $-\delta\lt x-1\lt\delta$ or $1-\delta\lt x\lt1+\delta$ Step 3. To show that $|f(x)-2|\geq1/2$, we need to find a $x$ so that $f(x)-2 \geq1/2$ or $2- f(x)\geq1/2$ which gives $f(x)\geq5/2$ or $f(x)\leq3/2$ Step 4 If $x\lt1$, we have $f(x)=x$, so we need $x\leq3/2$ for this case. If $x\gt1$, we have $f(x)=x+1$, so we need $x+1\geq5/2$ or $x+1\leq3/2$ which gives $x\geq3/2$ for this case. Step 5. As the results from step 3 and 4 overlap leaving no gap, there will always be a $x$ so that $|f(x)-2|\geq1/2$ which means that $\lim_{x\to1}f(x)\ne2$ (b) Repeat steps 3-5 above for $L=1$ instead of $2$. Step 3b. To show that $|f(x)-1|\geq1/2$, we need to find a $x$ so that $f(x)-1 \geq1/2$ or $1- f(x)\geq1/2$ which gives $f(x)\geq3/2$ or $f(x)\leq1/2$ Step 4b. If $x\lt1$, we have $f(x)=x$, so we need $x\leq1/2$ for this case. If $x\gt1$, we have $f(x)=x+1$, so we need $x+1\geq3/2$ or $x+1\leq1/2$ which gives $x\geq1/2$. Step 5b. As the results from step 3b and 4b overlap leaving no gap, there will always be a $x$ so that $|f(x)-1|\geq1/2$ which means that $\lim_{x\to1}f(x)\ne1$ (c) Repeat steps 3-5 above for $L=1.5$ instead of $2$. Step 3c. To show that $|f(x)-1.5|\geq1/2$, we need to find a $x$ so that $f(x)-1.5 \geq1/2$ or $1.5- f(x)\geq1/2$ which gives $f(x)\geq2$ or $f(x)\leq1$ Step 4c. If $x\lt1$, we have $f(x)=x$, so we need $x\leq1$ for this case. If $x\gt1$, we have $f(x)=x+1$, so we need $x+1\geq2$ or $x+1\leq1$ which gives $x\geq1$. Step 5c. As the results from step 3c and 4c overlap leaving no gap, there will always be a $x$ so that $|f(x)-1.5|\geq1/2$ which means that $\lim_{x\to1}f(x)\ne1.5$