Limit of a Piecewise Function

In this tutorial we shall discuss the limit of a piecewise function. Let us consider an example of limit of piecewise function.
For what value of a, \mathop {\lim }\limits_{x \to 2} f\left( x \right) exist, 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 left hand and right hand limits. For left hand limit x must approach 2 from left side, i.e. from the values less than 2 so 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 right hand limit x must approach 2 from right side, i.e. from the values greater than 2, so for 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}

Comments

comments