Answer:
[tex]\displaystyle\lim_{x \to \infty} \frac{x^2-1}{x^2+1} \\ \text{take $x^2$ common from numerator and denominator} \\ \displaystyle\lim_{x \to \infty} \dfrac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}} \\
\text{Now put the limit}
\\
\displaystyle\lim_{x \to \infty} \frac{1-0}{1+0}= 1
[/tex]
