Question

If the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{x^2-1}{x-1},& x\ne 1\\ k,& x=1\end{array}\right.$ is given to be continuous at x = 1, then the value of k is ____________.

Answer 1

The function $f\left(x\right)=\left\{\begin{array}{cc}\frac{x^2-1}{x-1},& x\ne 1\\ k,& x=1\end{array}\right.$ is continuous at x = 1.

$\therefore f\left(1\right)=\underset{x\to 1}{\lim }f\left(x\right)$

$\Rightarrow k=\underset{x\to 1}{\lim }\frac{x^2-1}{x-1}$

$\Rightarrow k=\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x+1\right)}{x-1}$

$\Rightarrow k=\underset{x\to 1}{\lim }\left(x+1\right)$

$\Rightarrow k=1+1=2$

Thus, the value of k is 2.


If the function $f\left(x\right)=\left\{\begin{array}{cc}\frac{x^2-1}{x-1},& x\ne 1\\ k,& x=1\end{array}\right.$ is given to be continuous at x = 1, then the value of k is ___2___.