Question

求极限 (x^2+16)/(16-x^2),其中x→-4

Answer 1

令t=4+x x=t-4

x→-4

t→0

x=t-4 代入得

lim(x^2+16)/(16-x^2) x→-4

=lim(t^2-8t+32)/(8t-t^2) t→0

=-1+（32)/(8t-t^2)

t→0时,8t-t^2→0,（32)/(8t-t^2)→0无穷

所以无极限