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Question
For what value of k is the following function continuous at x = 2?
Solution
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
\[\Rightarrow 5 = 5 = k\]
Hence, for k = 5,
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