## Calculus 8th Edition

sample answer: $\delta=0.01$
Graph (see below) y=$\tan$x, $y=1\pm 0.2$ (parallel lines, 0.2 above and below y=1 , to outline the interval on the y-axis to which the function values belong), $x=\arctan 0.8,\quad x=\arctan 1.2$ vertical lines to outline the interval $I=(0.675, 0.876)$ around x=$\displaystyle \frac{\pi}{4}$. From the graph, $x\in I \ \ \Rightarrow\ \ |f(x)-1| < 0.2$. (all values of $x\in I$ have f(x) within $\pm 0.2$ of $1$) We take a subset from $I$, one that we can write as$(\displaystyle \frac{\pi}{4}-\delta,\frac{\pi}{4}+\delta)$, such as, for example, for $\delta=0.01:$ $I_{1}=(\displaystyle \frac{\pi}{4}-0.01, \frac{\pi}{4}+0.01)$ (the problem asks to find a $\delta$, not the greatest possible $\delta)$ This interval is such that, for $x\in I_{1}\ \ \Rightarrow\ \ x\in I\ \ \Rightarrow\ \ |f(x)-1| < 0.2$. that is, $0 < |x-\displaystyle \frac{\pi}{4}| < \delta$, then $|\tan \mathrm{x}-1| < \epsilon$.