Let
f(x)=\begin{cases}\frac{x^{2}-16}{x-4}~~ , if ~~x \neq 4\\kx-24,~~if ~~~ x=4 \end{cases}
Step By Step Solution:
The given piecewise-defined function is:
f(x)=\begin{cases}\frac{x^{2}-16}{x-4}~~ , if ~~x \neq 4\\kx-24,~~if ~~~ x=4 \end{cases}
Notice that,
f(x) =\frac{x^{2}-16}{x-4}is not defined at x=4.But,
\lim_{x \to 4}f(x)is defined at x=4.Now,
f(x) =kx-24
is defined at x=4:
Using the Continuity of a Function,
\lim_{x \to a}f(x)=f(a)\lim_{x \to 4}f(x)=f(4)
Simplify the expression for x=4:
f(x)=kx-24~~, at ~x=4 \\f(4)=4k-24
As we know,
\lim_{x \to 4}f(x)=f(4)
Substitute the
f(x)=\frac{x^{2}-16}{x-4} \\and \\f(4)=4k-24into the above ,we get:
\lim_{x \to 4}{\left(\frac{x^{2}-16}{x-4}\right)}=4k-24
Evaluate the limit:
\lim_{x \to 4}{\left(\frac{x^{2}-16}{x-4}\right)}=4k-24\implies\lim_{x \to 4}{\left(\frac{\cancel{(x-4)}(x+4)}{\cancel{x-4}}\right)}=4k-24~~~~~~~~~~~~{\left (a^2-b^2=(a-b)(a+b) \right)}\implies\lim_{x \to 4}{\left(x+4\right)}=4k-24Now , putting values as discontinuity is removed:
\implies8=4k-24
\implies4k=32
\therefore k=8