Question

For what value of k is the following function continuous at x = 1? \[f\left( x \right) = \begin{cases}\frac{x^2 - 1}{x - 1}, & x \neq 1 \\ k , & x = 1\end{cases}\]

Answer 1

Given: 

\[f\left( x \right) = \binom{\frac{x^2 - 1}{x - 1}, x \neq 1}{k, x = 1}\]

If

\[f\left( x \right)\] is continuous at x = 1, then

\[\lim_{x \to 1} f\left( x \right) = f\left( 1 \right)\]

\[\lim_{x \to 1} \frac{x^2 - 1}{x - 1} = k\]

\[\lim_{x \to 1} \frac{\left( x - 1 \right)\left( x + 1 \right)}{x - 1} = k\]

\[\lim_{x \to 1} \left( x + 1 \right) = k\]

\[k = 2\]