## Thomas' Calculus 13th Edition

See graph; when $x\to0^+, y\to\infty$ or $\lim_{x\to 0^+}\frac{x^2+1}{x^2}=\infty$; while when $x\to0^-, y\to\infty$ or $\lim_{x\to 0^-}\frac{x^2+1}{x^2}=\infty$; the asymptote of the function is $x=0$