Question

For what value of k is the function $f\left(x\right)=\{\begin{array}{c}\frac{x^3+1}{x+1}x\ne -1\\ 2kx=-1\end{array}$ continuous at $x=2$

Answer 1

Given $f\left(x\right)$ is continuous at $x=2$

$\therefore {lim}_{x\to 2}f\left(x\right)=2k$

${lim}_{x\to 2}\frac{x^3+1}{x+1}=2k$

${lim}_{x\to 2}\frac{(x+1)(x^2+x+1)}{x+1}=2k$

${lim}_{x\to 2}{x}^2+x+1=2k$

$\Rightarrow 2^2+2+1=2k$

$\Rightarrow 2k=4+2+1=7$

$\therefore k=\frac{7}{2}$