A function has a vertical asymptote [tex]x=a[/tex] at point a, where the denominator becomes equal to 0.
A. The denominator of the function [tex]f(x)=\dfrac{x}{1-x^2}[/tex] turns into 0 at [tex]x=1[/tex] or [tex]x=-1.[/tex] Then [tex]x=1[/tex] and [tex]x=-1[/tex] are two vertical asymptotes of this function.
B. The denominator of the function [tex]f(x)=\dfrac{5x}{1-2x^2}[/tex] turns into 0 at [tex]x=\sqrt{\frac{1}{2}}[/tex] or [tex]x=-\sqrt{\frac{1}{2}}[/tex] Then [tex]x=\sqrt{\frac{1}{2}}[/tex] and [tex]x=-\sqrt{\frac{1}{2}}[/tex] are two vertical asymptotes of this function.
C. The denominator of the function [tex]f(x)=\dfrac{5x-1}{3+x^2}[/tex] never turns into 0, then this function hasn't any asymptotes.
D. The denominator of the function [tex]f(x)=\dfrac{x}{x+x^2}[/tex] turns into 0 at [tex]x=0.[/tex] Then [tex]x=0[/tex] is vertical asymptote of this function.
Answer: correct choice is C.
