Question

If \[f\left( x \right) = \begin{cases}\frac{x^2 - 16}{x - 4}, & \text{ if }  x \neq 4 \\ k , & \text{ if }  x = 4\end{cases}\]  is continuous at x = 4, find k.

Answer 1

Given: \[f\left( x \right) = \begin{cases}\frac{x^2 - 16}{x - 4}, & \text{ if }  x \neq 4 \\ k , & \text{ if }  x = 4\end{cases}\]

If  \[f\left( x \right)\]  is continuous at  \[x = 4\] , then

\[\lim_{x \to 4} f\left( x \right) = f\left( 4 \right)\]

\[\Rightarrow \lim_{x \to 4} \left( \frac{x^2 - 16}{x - 4} \right) = k\]

\[\Rightarrow \lim_{x \to 4} \frac{\left( x + 4 \right)\left( x - 4 \right)}{\left( x - 4 \right)} = k\]
\[ \Rightarrow \lim_{x \to 4} \left( x + 4 \right) = k\]
\[ \Rightarrow k = 8\]