Question

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

Answer 1

Hint: 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.