Chapter 8, Problem 2P

To determine

To find: The ${\lim }_{n\to \infty }\angle P_{n}A{P}_{n+1}$ .

Answer to Problem 2P

The value of limit ${\lim }_{n\to \infty }\angle P_{n}A{P}_{n+1}$ is $\frac{\pi }{3}$ .

Explanation of Solution

Given:

It is given that $\left|A{P}_1\right|=1$ , $\left|P_n{P}_{n+1}\right|=2^{n-1}$ , and $A{P}_n{P}_{n+1}$ is a right angle.

Calculation:

Find the length of $A{P}_2$ by use of Pythagoras Theorem.

  $\begin{array}{c}{\left|A{P}_2\right|}^2=A{P}_1^2+P_1{P}_2^2\\ =1^2+1^2\\ =2\end{array}$

Find the length of $A{P}_3$ by use of Pythagoras Theorem.

  $\begin{array}{c}{\left|A{P}_3\right|}^2=A{P}_2^2+P_2{P}_3^2\\ ={\left(\sqrt{2}\right)}^2+2^2\\ =2+2^2\end{array}$

Find the length of $A{P}_4$ by use of Pythagoras Theorem.

  $\begin{array}{c}{\left|A{P}_4\right|}^2=A{P}_3^2+P_3{P}_4^2\\ ={\left(\sqrt{2+2^2}\right)}^2+4^2\\ =2+2^2+{\left(2^2\right)}^2\end{array}$

Find the length of $A{P}_5$ by use of Pythagoras Theorem.

  $\begin{array}{c}{\left|A{P}_5\right|}^2=A{P}_4^2+P_4{P}_5^2\\ ={\left(\sqrt{2+2^2+{\left(2^2\right)}^2}\right)}^2+8^2\\ =2+2^2+{\left(2^2\right)}^2+{\left(2^3\right)}^2\end{array}$

Write the nth term of the series by mathematical induction.

  $\begin{array}{c}{\left|A{P}_n\right|}^2=2+2^2+{\left(2^2\right)}^2+\mathrm{...}+{\left(2^{n-2}\right)}^2\\ =2+\left(4+4^2+4^3+\mathrm{...}+4^{n-2}\right)\\ =2+\frac{4\left(4^{n-2}-1\right)}{4-1}\\ =\frac{2+4^{n-1}}{3}\end{array}$

Find the value of $\tan \left(\angle P_{n}A{P}_{n+1}\right)$ .

  $\begin{array}{c}\tan \left(\angle P_{n}A{P}_{n+1}\right)=\frac{\left|P_n{P}_{n+1}\right|}{\left|A{P}_n\right|}\\ =\frac{2^{n-1}}{\sqrt{\frac{2+4^{n-1}}{3}}}\\ =\frac{\sqrt{4^{n-1}}}{\sqrt{\frac{2+4^{n-1}}{3}}}\\ =\frac{1}{\sqrt{\frac{2}{3\cdot 4^{n-1}}+\frac{1}{3}}}\end{array}$

Apply the limit in the above equation as follows.

  $\begin{array}{c}\tan \left(\angle P_{n}A{P}_{n+1}\right)=\lim \left(\frac{1}{\sqrt{\frac{2}{3\cdot 4^{n-1}}+\frac{1}{3}}}\right)\\ =\sqrt{3}\\ \angle P_{n}A{P}_{n+1}={\tan }^{-1}\left(\sqrt{3}\right)\\ =\frac{\pi }{3}\end{array}$

Therefore, the value of limit ${\lim }_{n\to \infty }\angle P_{n}A{P}_{n+1}$ is $\frac{\pi }{3}$ .