If the function

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

$\therefore f(1)=\lim _{x \rightarrow 1} f(x)$

$\Rightarrow k=\lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1}$

$\Rightarrow k=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{x-1}$

$\Rightarrow k=\lim _{x \rightarrow 1}(x+1)$

$\Rightarrow k=1+1=2$

Thus, the value of k is 2.

If the function $f(x)=\left\{\begin{array}{cl}\frac{x^{2}-1}{x-1}, & x \neq 1 \\ k, & x=1\end{array}\right.$ is given to be continuous at $x=1$, then the value of $k$ i___2___.