Answer:
The function is continuous for all real numbers
Step-by-step explanation:
We have the following function
[tex]y=\frac{x^2}{x^2+1}[/tex]
Note that the denominator of the function is:
[tex]x^2 +1[/tex]
This expression is different from zero for all real numbers, since for [tex]x^2 +1[/tex] then [tex]x^2 =-1[/tex], there is no number in the real numbers whose square root is equal to -1.
For this reason the function is defined for all real numbers and has no discontinuity.
This function is always positive, continuous and has horizontal asymptote
[tex]y = 1[/tex]
Observe the attached image
