cho hàm số y=(x+căn (x+2017))/căn(x^2-1) có bao nhiêu đường tiệm cận

$$\eqalign{ & y = {{x + \sqrt {x + 2017} } \over {\sqrt {{x^2} - 1} }} \cr & DKXD:\,\,\left\{ \matrix{ {x^2} - 1 > 0 \hfill \cr x + 2017 \ge 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x > 1 \hfill \cr x < - 1 \hfill \cr} \right. \hfill \cr x \ge - 2017 \hfill \cr} \right. \cr & \Rightarrow D = \left[ { - 2017;1} \right) \cup \left( {1; + \infty } \right) \cr & \mathop {\lim }\limits_{x \to {1^ + }} {{x + \sqrt {x + 2017} } \over {\sqrt {{x^2} - 1} }} = + \infty \cr & \mathop {\lim }\limits_{x \to - {1^ - }} {{x + \sqrt {x + 2017} } \over {\sqrt {{x^2} - 1} }} = + \infty \cr & Vay\,\,do\,thi\,ham\,\,so\,\,co\,\,2\,\,TCD\,\,x = \pm 1 \cr} $$ $$\mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{x \to + \infty } {{x + \sqrt {x + 2017} } \over {\sqrt {{x^2} - 1} }} = \mathop {\lim }\limits_{x \to + \infty } {{1 + \sqrt {{1 \over x} + {{2017} \over {{x^2}}}} } \over {\sqrt {1 - {1 \over {{x^2}}}} }} = 1$$ => ĐTHS có TCN y=1