Given:
[tex]\lim _{x\to4}\frac{x^3-64}{x-4}[/tex]
solve:
[tex]\begin{gathered} \lim _{x\to4}\frac{(x)^3-64}{x-4} \\ =\lim _{x\to4}\frac{(x)^3-(4)^3}{x-4} \\ =\lim _{x\to4}\frac{(x-4)(x^2+4x+4^2)}{x-4} \\ =\lim _{x\to4}(x^2+4x+16) \end{gathered}[/tex][tex]\begin{gathered} =\lim _{x\to4}(x^2+4x+16) \\ =4^2+4(4)+16 \\ =16+16+16 \\ =48 \end{gathered}[/tex]
The value of function is 48 and and extimate limit is 16.
