The lmit of arctan(x) as x approaches infinity is examined using two different approaches. The first one is based on the right triangle and the second is based of the inverse function arctan(x) definition.
Arctan(x) in a Right Triangle Approach
Let ? = arctan(x) which gives tan ? = x = x / 1 and use the definition of tan x in a right triangle to visualise x = tan ? .
Fig. 1 - ? = arctan(x) in a Right Triangle Geometrical Approach
As x increases indefinitely, geometrically ? approaches ?/2 and hence we write the limit
Algebraic Approach
We can also use sin (?) given by
Calculate the limit
\( \) \( \)\( \)\( \)
which gives
\[ \lim_{x \to \infty} \sin \alpha = \sin \dfrac{\pi}{2} \]
Using the limit of composite functions, we write \( \lim_{x \to \infty} \sin \alpha = \sin (\lim_{x \to \infty} \alpha) \) and the fact that
\( \dfrac{\pi}{2} \) is a constant hence \( \dfrac{\pi}{2} = \lim_{x \to \infty} \dfrac{\pi}{2} \), we write
\[ \lim_{x \to \infty} \sin \alpha = \sin (\lim_{x \to \infty} \; \alpha ) = \sin \lim_{x \to \infty} \dfrac{\pi}{2} \]
Using \( \alpha = \arctan(x) \) and the above, we write
\[ \lim_{x \to \infty} \; \alpha = \lim_{x \to \infty} \; \arctan(x) = \dfrac{\pi}{2} \]
Tan (x) and Arctan(x) Graphs Approach
Below is shown the graph of the function function \( y = \tan x \) (in blue) on the interval \( (-\dfrac{\pi}{2} , \dfrac{\pi}{2}) \), with \( x = \dfrac{\pi}{2} \) and \( x = - \dfrac{\pi}{2} \) being vertical asymptotes (broken line blue). It is a one to one function function and therefore has the inverse function \( y = \arctan (x) \) as shown below (in red).
The vertical asymptotes of \( y = \tan x \) becomes horizontal asymptotes (broken line red) of the inverse function \( y = \arctan (x) \) which by definition may be written using limits as follows:
\[ \lim_{x \to \infty} \arctan(x) = \dfrac{\pi}{2} \]
and
\[ \lim_{x \to -\infty} \arctan(x) = -\dfrac{\pi}{2} \]
Fig. 2 - Graph of \( \tan(x) \) and \( \arctan(x) \)
