`y = sec^2(x)/(x^2 + 1)` Find the limit, if possible

Expert Answers
gsarora17 eNotes educator| Certified Educator

`lim_(x->oo)sec^2(x/(x^2+1))`

`lim_(x->oo)sec^2(x)/(x^2(1+1/x^2))`

`lim_(x->oo)sec^2(1/(x(1+1/x^2)))`

plug in the value of x to evaluate the limit

`= sec^2(1/(oo(1+0)))`

`= sec^2(0)=`

`lim_(x->0)sec^2(x/(x^2+1))`

plug in the value of x=0

`= sec^2(0/(0+1))`

`= sec^2(0)=1`