Question

How do you find the limit of $arctan\left(x\right)$ as $x$ approaches infinity?

Answer 1

Hint: The question is solved by finding out the principal value of the function by putting $x=\mathrm{\infty }$ . Then the general limit of the inverse tan function is obtained which is the sum of principal value and $n\pi$ . As the values of x are tending to positive infinity only the specific range of the general limit is considered.

Complete step by step solution:
The $\arctan \left(x\right)$ is the inverse of the $\tan x$ function. It returns an angle for the numerical value of the tangent given.
$\Rightarrow Arctan\left(x\right)={\displaystyle \frac{1}{tan\left(x\right)}}$
The domain of the trigonometric function is the set of all real numbers and the function ranges from $-\dfrac{\pi }{2}The graph of the inverse tangent function is symmetrical about the x-axis and is bounded between $-{\displaystyle \frac{\pi }{2}}$ and ${\displaystyle \frac{\pi }{2}}$
The graph neither has any maxima or minima having a period of $\pi$
The general solution or principal values of the inverse tangent function is given as,
 $\Rightarrow 2\pi k+{\displaystyle \frac{\pi }{2}}$ ; as $k$ approaches positive infinity.
The above equation can be simplified as,
$\Rightarrow \pi (2k+{\displaystyle \frac{1}{2}})$
Let us assume that $y=arctan\left(x\right)$
According to the question,
From the trigonometric table, we know that $arctan\left(\mathrm{\infty }\right)$ is ${\displaystyle \frac{\pi }{2}}$
Writing the same we get,
${\displaystyle \underset{x\to {\displaystyle \frac{\pi }{2}}}{lim}tanx=+\mathrm{\infty }}$
Now taking the tangent on the other side,
$\Rightarrow {\displaystyle \underset{x\to +\mathrm{\infty }}{lim}y={\displaystyle \frac{\pi }{2}}}$
As the value of x tends to infinity the principal value of y obtained is ${\displaystyle \frac{\pi }{2}}$
The general limit for the above trigonometric function is given as the sum of principal value and kπ where $k=0,\pm 1,\pm 2.....$
Hence the values of x approaching positive infinity the values of k must be even which means the general limit is equal to ${\displaystyle \frac{\pi }{2}}+2k\pi$ where $k=0,\pm 1,\pm 2.....$.

Note: We must know the formula for the general solutions of $arctan\left(x\right)$ to solve the question in an easy way. We must note that the general limit of the function is given by the principal value $+n\pi$. The general limit of the values of x approaching negative infinity is given by ${\displaystyle \frac{\pi }{2}}+(2k+1){\displaystyle \frac{\pi }{2}}$ where $k=0,\pm 1,\pm 2.....$ that is only the odd integral of $\pi$.