Question

How do you find the limit of $\dfrac{{|x - 3|}}{{x - 3}}$ as $x$ approaches ${3^ - }?$

Answer 1

In the question we have to find the limit of a function such that $x$ approaches ${3^ - }$ . Here first we need to understand what ${3^ - }$ means. When we say $x \to {3^ - }$ it means that the value of $x$ approaches $3$ from the left-hand side in the number system. In order to solve this question, we will use the formula for left-hand limit i.e., $\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {a - h} \right)$ . After that we will simplify and get the result.

Complete step by step answer:
We are given the function $f\left( x \right) = \dfrac{{|x - 3|}}{{x - 3}}$
And we have to calculate the value of $\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}}$
Now using the formula of the left-hand limit of a function i.e.,
$\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {a - h} \right)$
Here, $a = 3$
Therefore, we have
$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {3 - h} \right){\text{ }} - - - \left( i \right)$
Since we have $f\left( x \right) = \dfrac{{|x - 3|}}{{x - 3}}$
Then the value of $f\left( {3 - h} \right) = \dfrac{{|3 - h - 3|}}{{3 - h - 3}}$
Now on substituting in the equation $\left( 1 \right)$ we get
$\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{|3 - h - 3|}}{{3 - h - 3}}$
On solving right-hand side of the above expression, we get
$\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{| - h|}}{{ - h}}$
Now we know that
$| - x|{\text{ }} = x$
Therefore, we get
$\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}} = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{ - h}}$
On cancelling numerator and denominator, we have
$\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}} = \mathop {\lim }\limits_{h \to 0} {\text{ }} - 1$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}} = - 1$
Hence, the value of the function $\dfrac{{|x - 3|}}{{x - 3}}$ as $x$ approaches ${3^ - }$ is $- 1$

Note:
Limit of a function $f\left( x \right)$ such that $x$ approaches ${a^ - }$ is called the left-hand limit of the function as it approaches from the left-hand side. It also means that $x$ is some value which is less than $3$ and is moving towards the rightward direction in the number system line.
Also note there is an alternative way to solve this problem such as:
We are given the function $f\left( x \right) = \dfrac{{|x - 3|}}{{x - 3}}$
And we have to calculate the value of $\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}}$
As $x$ is some value which is less than $3$
Therefore, $|x - 3|{\text{ }} = - \left( {x - 3} \right)$
So, we get
$\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}}{\text{ }} = {\text{ }}\mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{ - \left( {x - 3} \right)}}{{x - 3}}$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}}{\text{ }} = {\text{ }}\mathop {\lim }\limits_{x \to {3^ - }} - 1$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} \dfrac{{|x - 3|}}{{x - 3}}{\text{ }} = - 1$
which is the required answer.