Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 9

Find the values of $\delta$ that corresponds to (a)$ M = 1000$ and (b)$ M = 10,000$ for the $\lim \limits_{x \to \frac{\pi}{2}} \tan^2 x = \infty$

First, we will get the values of $x$ that intersect at the given curve and the line $y=1000$. Let $x_L$ and $x_R$
are the values of $x$ from the left and right of $\displaystyle \frac{\pi}{2}$ respectively.








$
\begin{equation}
\begin{aligned}

\tan^2 x =& 1000\\

\sqrt{\tan^2x} =& \sqrt{1000} = 10\sqrt{10}\\

x =& \tan^{-1 } [10\sqrt{10}]\\

x_L =& 1.5392 \text{ and } x_R = 1.602

\end{aligned}
\end{equation}
$


Now, we can determine the value of $\delta$ by checking the values of $x$ that would give a smaller distance to $\displaystyle \frac{\pi}{2}$.


$
\begin{equation}
\begin{aligned}
\frac{\pi}{2} - x_L = \frac{\pi}{2} - 1.5392 & = 0.0316\\
\frac{\pi}{2} - x_R = 1.602 - \frac{\pi}{2} & = 0.0312
\end{aligned}
\end{equation}
$


Hence,

$\quad \fbox{$\delta \leq 0.0312$}$

Using the same method above for $M = 10,000$ we get...





$
\begin{equation}
\begin{aligned}

\tan^2 x =& 10000\\

\sqrt{\tan^2x} =& \sqrt{10000} = 100\\

x =& \tan^{-1 } [100]\\

x_L =& 1.5608 \text{ and } x_R = 1.581

\end{aligned}
\end{equation}
$

Now, we can determine the value of $\delta$ by checking the values of $x$ that would give a smaller distance to $\displaystyle \frac{\pi}{2}$.


$
\begin{equation}
\begin{aligned}
\frac{\pi}{2} - x_L = \frac{\pi}{2} - 1.5608 & = 0.0099\\
\frac{\pi}{2} - x_R = 1.581- \frac{\pi}{2} & = 0.0102
\end{aligned}
\end{equation}
$


Hence,

$\quad \fbox{$\delta \leq 0.0099$}$