Question

How do you find the limit of $\arctan (x) $ as $x$ approaches to $\infty $?

Answer 1

Hint:In order to determine the above limit ,we must consider the fact the range of the function $\arctan (x)$ is the angle in $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$and we know that $\tan 0 = 0$,so if we increase the value of $\theta $,the tangent also increases approaches the value infinity $\infty $. The angle at which the tangent increments without having any bound and approaches the value infinity $\infty $ is $\dfrac{\pi }{2}$.

Complete step by step solution:
Recall the fact that the range of the function $\arctan (x)$ is the angle \[\theta \] in the interval of $\left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$having $\tan \theta = x$
To address the question, we need to sort out:
What does $\theta $ need to draw near for the tangent to get more greater and greater with no bound on how big it gets?
Note that $\tan 0 = 0$, however $\theta $ as increments, so does the tangent.
Truth be told as $\theta $ gets nearer to $\dfrac{\pi }{2}$ the tangent increments without bound. So on the off chance that we increment the tangent without bound, at that point the corresponding angle (number) approaches $\dfrac{\pi }{2}$
Therefore , $ \mathop {\lim }\limits_{x \to \infty } \,\arctan x = \dfrac{\pi }{2}$.

Note:
1.Limit: You and your companions choose to meet at some spot outside. Is it essential that every one of your companions is living in a similar spot and stroll on a similar street to arrive at that place?
Actually no, not generally. All companions come from various pieces of the city or nation to meet at that one single spot.
It would appear that intermingling of various components to a solitary point. Mathematically, it resembles an intermingling of a function to a specific value. It is an illustration of cutoff points. Cut- off points show how a few functions are limited. The function watches out for some worth when its breaking point moves toward some value.
2. Don’t forget to cross-check your answer.