Question

How do you find one sided limits and limits of piecewise functions?

Answer 1

Hint:To find one sided limits, we have to find the limit either from left hand side or right hand side depending on the function. One sided limits exist when anyone of the left hand limit or the right hand limit exists and equals the functional value at that point. In order to find limits of piecewise function, we have to take one sided limits separately, as a piecewise function has different formulas or functions in different intervals.

Complete step by step solution:
We will first know how to find one sided limits, let us understand it with an example
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{0,}&{{\text{if}}\;x < 1} \\
{1,}&{{\text{if}}\;x \geqslant 1}
\end{array}} \right\}\]
In this above function we approach $1$ from left hand side then $f(x) = 0$ and if we approach $1$ from right hand side then $f(x) = 1$
$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} = 0,\;\mathop {\lim }\limits_{x \to {1^ + }} = 1\;{\text{and}}\;f(1) = 1$
In this example, one sided (right hand limit) limit exists, because the value of right hand limit and the functional value is equal.
Now coming to limits for piecewise function,
For the following piecewise defined function,
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
{2x,}&{{\text{if}}\;x < 2} \\
{x^2}, \\
x - 1, \\
&
{\text{if}}\;3 \geqslant x \geqslant 2 \\
{\text{if}}\;x > 3 \\
\end{array}} \right\}\]
We will check here the limits separately for $\mathop {\lim }\limits_{x \to 2} \;{\text{and}}\;\mathop
{\lim }\limits_{x \to 3} $
a) $\mathop {\lim }\limits_{x \to 2} f(x)$
We have to check here both one sided limits separately, because function $f(x)$ is different for left and right hand limit of $2$
$
\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} (2x) = 2
\times 2 = 4 \\
\mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} ({x^2}) =
{2^2} = 4 \\
$
Therefore, we got the value $\mathop {\lim }\limits_{x \to 2} f(x) = 4$
b) $\mathop {\lim f(x)}\limits_{x \to 3} $
Similarly, checking both side limits for $\mathop {\lim }\limits_{x \to 3} $
$
\mathop {\lim }\limits_{x \to {3^ - }} f(x) = \mathop {\lim }\limits_{x \to {3^ - }} ({x^2}) =
{3^2} = 9 \\
\mathop {\lim }\limits_{x \to {3^ + }} f(x) = \mathop {\lim }\limits_{x \to {3^ + }} (x - 1) = 3 - 1
= 2 \\
$
Here $\mathop {\lim f(x)}\limits_{x \to {3^ - }} \ne \mathop {\lim }\limits_{x \to {3^ + }} f(x)$
therefore \[\mathop {\lim }\limits_{x \to 3} f(x)\] does not exist.

Note: We can get one sided limits even when the normal limit doesn’t exist. If both one sided limits don’t have equal value, then the normal limit does not exist. The only difference between one sided limit and normal limit is the range of $x's$ when finding the values. Piecewise functions should be solved by making cases of each limit in order to make the process easy.