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 $x=\infty$ . 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.
$\Rightarrow Arctan\left( x \right)\text{ }=\text{ }\dfrac{1}{tan\left( x \right)}$
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 $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$
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,
 $\Rightarrow 2\pi k~+~\dfrac{\pi }{2}$ ; as $k$ approaches positive infinity.
The above equation can be simplified as,
$\Rightarrow \pi \left( 2k\text{ }+\text{ }\dfrac{1}{2} \right)$
Let us assume that $y\text{ }=\text{ }arctan\left( x \right)$
According to the question,
From the trigonometric table, we know that $arctan\left( \infty \right)$ is $\dfrac{\pi }{2}$
Writing the same we get,
$\displaystyle \lim_{x \to \dfrac{\pi }{2}}tanx=+\infty$
Now taking the tangent on the other side,
$\Rightarrow \displaystyle \lim_{x \to +\infty }y=\dfrac{\pi }{2}$
As the value of x tends to infinity the principal value of y obtained is $\dfrac{\pi }{2}$
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 $\dfrac{\pi }{2}\text{ }+\text{ }2k\pi$ 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 $\dfrac{\pi }{2}\text{ }+\text{ }\left( 2k+1 \right)\dfrac{\pi }{2}$ where $k=0,\pm 1,\pm 2.....$ that is only the odd integral of $\pi$.