Đáp án:
TCN: y=1 vaf y=-1.
Giải thích các bước giải:
\(\eqalign{
& y = {{x + 3} \over {\sqrt {{x^2} + 1} }} \cr
& \mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } {{x + 3} \over {\sqrt {{x^2} + 1} }} = 1 \cr
& \mathop {\lim }\limits_{x \to - \infty } y = \mathop {\lim }\limits_{x \to - \infty } {{x + 3} \over {\sqrt {{x^2} + 1} }} = - 1 \cr
& \Rightarrow Ham\,\,so\,\,co\,\,TCN\,\,y = \pm 1. \cr
& Ham\,\,so\,\,co\,\,TXD\,\,D = R \Rightarrow DTHS\,\,ko\,\,co\,\,TCD. \cr} \)
