As you have #x^2# then #1+x^2# will always be positive. So #y# is always positive.
As #x# becomes smaller and smaller then #1/(1+x^2) -> 1/1 = 1#
So #lim_(x->0) 1/(1+x^2)=1#
As #x# becomes bigger and bigger then #1+x^2# becomes bigger so #1/(1+x^2)# becomes smaller.
#lim_(x->+-oo) 1/(1+x^2)=0#
#color(blue)("build a table of value for different values of "x" and calculate the appropriate values of y. Plot")##color(blue)(" the obtained values of y against those chosen for x")#