Question

For what value of k is the following function continuous at x = 2? 

\[f\left( x \right) = \begin{cases}2x + 1 ; & \text{ if } x < 2 \\ k ; & x = 2 \\ 3x - 1 ; & x > 2\end{cases}\]

Answer 1

Given: 

\[f\left( x \right) = \begin{cases}2x + 1 ; & \text{ if } x < 2 \\ k ; & x = 2 \\ 3x - 1 ; & x > 2\end{cases}\]

We have
(LHL at x = 2) = 

\[\lim_{x \to 2^-} f\left( x \right) = \lim_{h \to 0} f\left( 2 - h \right)\]

\[= \lim_{h \to 0} \left( 2\left( 2 - h \right) + 1 \right) = 5\]

(RHL at x = 2) = 

\[\lim_{x \to 2^+} f\left( x \right) = \lim_{h \to 0} f\left( 2 + h \right)\]

\[= \lim_{h \to 0} 3\left( 2 + h \right) - 1 = 5\]

Also, 

\[f\left( 2 \right) = k\]

If f(x) is continuous at x = 2, then
​

\[\lim_{x \to 2^-} f\left( x \right) = \lim_{x \to 2^+} f\left( x \right) = f\left( 2 \right)\]

\[\Rightarrow 5 = 5 = k\]

Hence, for k = 5,

\[f\left( x \right)\]  is continuous at

\[x = 2\]