Question

If, is continuous at $x=2$, find the value of $k$.

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

Answer 1

Since the function $f\left(x\right)$ is continuous $x=2$ then $\underset{x\to 2}{lim}f\left(x\right)=f\left(2\right)$......(1).

Now,

$\underset{x\to 2}{lim}f\left(x\right)$

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

Using L'Hospital's rule we get,

$=\underset{x\to 2}{lim}\frac{3x^2+2x-16}{2(x-2)}$$\frac{0}{0}$ form

Again using L'Hospital's rule we get,

$=\underset{x\to 2}{lim}\frac{6x+2}{2}=7$.

From (1) we get, $k=7$.