Given the function:
[tex]k(x)=\frac{x^3-8x-32}{x-4}[/tex]
We need to find the value of k(x) when x = 3.9 and x = 3.999 and x = 4.001
So, for each value of x we will substitute into the function k(x)
when x = 3.9
[tex]k(x)=\frac{3.9^3-8\cdot3.9-32}{3.9-4}=\frac{-3.881}{-0.1}=38.81[/tex]
When x = 3.999
[tex]k(x)=\frac{3.999^3-8\cdot3.999-32}{3.999-4}=\frac{-0.039988}{-0.001}=39.9880012[/tex]
When x = 4.001
[tex]k(x)=\frac{4.001^3-8\cdot4.001-32}{4.001-4}=\frac{0.040012}{0.001}=40.0120007[/tex]
When x = 4.1
[tex]k(x)=\frac{4.1^3-8\cdot4.1-32}{4.1-4}=\frac{4.121}{0.1}=41.21[/tex]
So, we can deduce that The limit of the function = 40
