Question

Find the left hand limit and right hand limit of the function$f(x)={\displaystyle \frac{|x-2|}{x-2}}$ at $x=2$

Answer 1

Hint: Define the function for the values of $x<2$and$x>2$. Compute their limits using the properties of absolute value and the fact that the limit of a constant function is the constant.
These limits will be the required answer.

Complete step by step answer:
Finding the left hand and right hand limits of a function$f(x)$at a point$x=a$means finding the limit of $f(x)$ at $x<a$and finding the limit of $f(x)$ at $x>a$respectively where $a$ is any real number.
The left hand limit of $f(x)$ at $x<a$ is denoted by$\underset{x\to a^{-}}{lim}f(x)$ if it exists.
Similarly, the right hand limit of$f(x)$ at $x>a$is denoted by $\underset{x\to a^{+}}{lim}f(x)$ if it exists.
Therefore, to find the left and right hand limits we need to define the value of$f(x)$ at $x>a$and at $x<a$ respectively.
In the given question, we have
$a=2$ and $f(x)={\displaystyle \frac{|x-2|}{x-2}}$ at $x=2$
Therefore, we will determine the value of $f(x)={\displaystyle \frac{|x-2|}{x-2}}$ at $x<2$ and $x>2$
Let’s recall the behaviour of the absolute value function.
For a real number $x$
If $x<0$, then $|x|=-x$ and
if $x>0$, then $|x|=x$
Consider the graph of the absolute value function.

Thus, when $x<2$,
$$\begin{gathered} x-2<0 \\ \Rightarrow |x-2|=-(x-2) \end{gathered}$$
Therefore, $f(x)={\displaystyle \frac{|x-2|}{x-2}}={\displaystyle \frac{-(x-2)}{x-2}}=-1$
Thus, $f(x)$ is a constant function when $x<2$
We know that the limit of a constant function is equal to the constant.
This implies that $$\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}(-1)=-1$$.
That is, the left hand limit of the given function $f(x)={\displaystyle \frac{|x-2|}{x-2}}$ at $x=2$ is $-1$
Similarly, we need to find the right hand limit of $f(x)={\displaystyle \frac{|x-2|}{x-2}}$ at $x=2$
Now, when $x>2$
$$\begin{gathered} x-2>0 \\ \Rightarrow |x-2|=x-2 \end{gathered}$$
Therefore, we have $f(x)={\displaystyle \frac{|x-2|}{x-2}}={\displaystyle \frac{x-2}{x-2}}=1$
Below is the graph of the function$f(x)={\displaystyle \frac{|x-2|}{x-2}}$

Thus, $f(x)$ is a constant function when $x>2$ as well.
Let us compute the right hand limit of $f(x)$.
$$\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}(1)=1$$.
Thus, the right hand limit of the given function is 1.
Hence the left hand and right hand limits of the function $f(x)={\displaystyle \frac{|x-2|}{x-2}}$ at $x=2$ are $-1$ and $1$ respectively.

Note: It is advisable to draw a graph of a function to understand the nature of the function for which the left hand and right hand limits are to be calculated. Functions are best understood with the help of graphs.