Question

Find the value of k that makes f(x) continuous at x = 4

Answer 1

Given:

The function is continuous at x = 4,

$f(x)=\begin{cases}\frac{2x^2-8x}{x-4},x\ne4 \\ k\text{ ,x= 4}\end{cases}$

As the given function is continous at x = 4,

$\begin{gathered} \lim _{x\to4}f(x)=k \\ \lim _{x\to4}(\frac{2x^2-8x}{x-4})=k \\ \lim _{x\to4}(\frac{2x(x-4)}{x-4})=k \\ \lim _{x\to4}(2x)=k \\ 2(4)=k \\ k=8 \end{gathered}$

Answer: k = 8