No L'Hopital's rule needed:
[tex]\displaystyle\lim_{x\to8}\frac{x^2-64}{x^2-8x}=\lim_{x\to8}\frac{(x+8)(x-8)}{x(x-8)}=\lim_{x\to8}\frac{x+8}x=\dfrac{16}8=2[/tex]
but we can still use it if only to verify the result:
[tex]\displaystyle\lim_{x\to8}\frac{x^2-64}{x^2-8x}\stackrel{\mathrm{LHR}}=\lim_{x\to8}\frac{2x}{2x-8}=\lim_{x\to8}\frac x{x-4}=\frac8{8-4}=2[/tex]
