How do we know this is true? I wasn't sure where pi/2 came from. The image contains a handwritten mathematical expression:\[\lim_{{x \to \infty}} \arctan x = \frac{\pi}{2}\]This expression states that as $ x $ approaches infinity, the limit of $\arctan x$, or the inverse tangent of $ x $, is $\frac{\pi}{2}$. This is a fundamental concept in calculus and trigonometry, illustrating the behavior of the arctangent function as its input grows indefinitely large. The output approaches the asymptote at $\frac{\pi}{2}$.There are no graphs or diagrams in the image.

The given function is arctanx we have to find limit at x→∞. Now the limit exist if limx→∞+arctanx =…

Answered: lim arctanx %=D7

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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How do we know this is true? I wasn't sure where pi/2 came from.

The image contains a handwritten mathematical expression:

\[
\lim_{{x \to \infty}} \arctan x = \frac{\pi}{2}
\]

This expression states that as $ x $ approaches infinity, the limit of $\arctan x$, or the inverse tangent of $ x $, is $\frac{\pi}{2}$. This is a fundamental concept in calculus and trigonometry, illustrating the behavior of the arctangent function as its input grows indefinitely large. The output approaches the asymptote at $\frac{\pi}{2}$.

There are no graphs or diagrams in the image.

Expert Solution

Step 1

The given function is $a r c \tan x$ we have to find limit at $x \to \infty$ .

Now the limit exist if $\lim_{x \to \infty^{+}} a r c \tan x = \lim_{x \to \infty^{-}} a r c \tan x$ and the value of the limit will be

$\lim_{x \to \infty} a r c \tan x = \lim_{x \to \infty^{+}} a r c \tan x = \lim_{x \to \infty^{-}} a r c \tan x$

We know,

$\tan \frac{π}{2} = \frac{\sin (\frac{π}{2})}{\cos (\frac{π}{2})} = \frac{1}{0} = \infty \Rightarrow \tan \frac{π}{2} = \infty$

$\Rightarrow \frac{π}{2} = \tan^{- 1} (\infty)$

$\Rightarrow \tan^{- 1} (\infty) = \frac{π}{2} \Rightarrow a r c \tan (\infty) = \frac{π}{2}$

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