Question

Find the value of k,so that the following function is continuous at x = 2.
$f\left(x\right)=\{\begin{array}{c}f\left(x\right)=\frac{x^3+x^2-16x+20}{{(x-2)}^2},x\ne 2\\ k,x=2\end{array}$

Answer 1

$f\left(x\right)=\{\begin{array}{c}f\left(x\right)=\frac{x^3+x^2-16x+20}{{(x-2)}^2},x\ne 2\\ k,x=2\end{array}{lt}_{x\to 2}f\left(x\right)=f\left(2\right)$

${lt}_{x\to 2}\frac{x^3+x^2-16x+20}{{(x-2)}^2}=\frac{0}{0}$ form

Applying L' Hospitals rule

${lt}_{x\to 2}\frac{3x^2-12x-16}{2(x-2)}=\frac{0}{0}$ form

Again applying L'Hospital's rule

${lt}_{x\to 2}\frac{6x+2}{2}=\frac{6\left(2\right)+2}{2}=\frac{14}{2}=7k=7$