Limit of a Piecewise Function

In this tutorial we shall discuss the limit of a piecewise function. Let us consider an example of the limit of a piecewise function.

For what value of $$a$$, $$\mathop {\lim }\limits_{x \to 2} f\left( x \right)$$ exists, where
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {2ax,\,\,\,\,\,\,\,\,\,\,x < 2} \\ {6 – 2ax,\,\,\,x > 2} \end{array}} \right.\]

The given function $$f$$ is split into two parts: one is defined for $$x < 2$$ and the other is defined for $$x > 2$$. So, we have to take the left hand and right hand limits. For the left hand limit $$x$$ must approach 2 from the left side, i.e. from the values less than 2, so for the left hand limit we shall use the function part $$2ax$$. Thus,
\[\begin{gathered} \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left( {2ax} \right) \\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = 2a\mathop {\lim }\limits_{x \to 2} \left( x \right) \\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = 2a\left( 2 \right) = 4a \\ \end{gathered} \]

For the right hand limit $$x$$ must approach 2 from the right side, i.e. from the values greater than 2, so for the right hand limit we shall use the function part $$6 – 2ax$$. Thus,
\[\begin{gathered} \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \left( {6 – 2ax} \right) \\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = \left[ {6 – 2a\left( 2 \right)} \right] \\ \Rightarrow \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 6 – 4a \\ \end{gathered} \]

It is given that the limit $$\mathop {\lim }\limits_{x \to 2} f\left( x \right)$$ exists, so the left and right hand limits must be the same, i.e.
\[\mathop {\lim }\limits_{x \to {2^ – }} f\left( x \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\]
\[\begin{gathered} \Rightarrow 4a = 6 – 4a \Rightarrow 8a = 6 \\ \Rightarrow a = \frac{3}{4} \\ \end{gathered} \]