Without resorting to L'Hopitâl's rule,
[tex]\displaystyle\lim_{x\to9}\frac{x-9}{x^2-81}=\lim_{x\to9}\frac{x-9}{(x-9)(x+9)}=\lim_{x\to9}\frac1{x+9}=\frac1{18}[/tex]
With the rule, we get the same result:
[tex]\displaystyle\lim_{x\to9}\frac{x-9}{x^2-81}=\lim_{x\to9}\frac1{2x}=\frac1{18}[/tex]
