Answer:
Third Option
Step-by-step explanation:
We know that the function [tex]tan(x)[/tex] is defined as [tex]y=\frac{sin(x)}{cos(x)}[/tex]. Since the denominator is [tex]cos(x)[/tex] then we know that [tex]cos(x)=0[/tex] when [tex]x=\frac{\pi}{2}[/tex]
We also know that the division by 0 is not defined. Therefore, the limit of [tex]y=tan(x)[/tex] when "x" tends to [tex]\frac{\pi}{2}[/tex] is infinite.
The function [tex]tan^{-1}(x)[/tex] is the inverse of [tex]tan(x)[/tex]
By definition, if we have a function f(x), its domain will be equal to the range of its inverse function [tex]f^{-1}(x)[/tex]. If [tex]f(3)=8[/tex], then [tex]f^{-1}(8)=3[/tex]
This also happens for the function [tex]tan^{-1}(x)[/tex]
If when [tex]x \to \frac{\pi}{2}, tan(x) \to \infty[/tex] then when [tex]x \to \infty, tan^{-1}x \to \frac{\pi}{2}[/tex]
Then, the answer is:
[tex]x \to \infty, f(x) \to \frac{\pi}{2}[/tex]
