Question

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

Answer 1

Hint: We use the concept of limits and solve the value as limit approaches \[3 + \]. When we have to find the limit of a function such that x approaches \[3 + \] then we move to the number 3 from the right hand side on the number line.
* Limit of a function f(x) such that x approaches \[a + \] is called the right hand limit of the function as it approaches from the right hand side. \[\mathop {\lim }\limits_{x \to a + } f(x) = \mathop {\lim }\limits_{h \to 0} f(a + h)\]

Complete step by step solution:
We are given the function \[f(x) = \dfrac{{x + 3}}{{x - 3}}\] … (1)
We have to calculate the value of \[\mathop {\lim }\limits_{x \to 3 + } \dfrac{{x + 3}}{{x - 3}}\]
Use the formula of the right hand limit of a function. We know in the right hand limit, the values approach from the right hand side of the number i.e. we take values just greater than the number.
Since \[\mathop {\lim }\limits_{x \to a + } f(x) = \mathop {\lim }\limits_{h \to 0} f(a + h)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3 + } f(x) = \mathop {\lim }\limits_{h \to 0} f(3 + h)\] … (2)
Since we have \[f(x) = \dfrac{{x + 3}}{{x - 3}}\]
Then the value of \[f(3 + h) = \dfrac{{(3 + h) + 3}}{{(3 + h) - 3}}\]
Add terms in numerator and denominator separately.
\[ \Rightarrow f(3 + h) = \dfrac{{3 + h + 3}}{{3 + h - 3}}\]
\[ \Rightarrow f(3 + h) = \dfrac{{h + 6}}{h}\]
Now separate the numerator into two fractions by taking the denominator with each term of the numerator.
\[ \Rightarrow f(3 + h) = \dfrac{h}{h} + \dfrac{6}{h}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow f(3 + h) = 1 + \dfrac{6}{h}\] … (3)
Substitute the value of the function from equation (3) in equation (2)
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3 + } \dfrac{{x + 3}}{{x - 3}} = \mathop {\lim }\limits_{h \to 0} (1 + \dfrac{6}{h})\]
Put the value of ‘h’ as 0 in the function
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3 + } \dfrac{{x + 3}}{{x - 3}} = (1 + \dfrac{6}{0})\]
We know that when \[h \to 0\]then \[\dfrac{6}{h} \to \infty \]
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3 + } \dfrac{{x + 3}}{{x - 3}} = (1 + \infty )\]
Also, we know any value added to infinity results in infinity.
\[ \Rightarrow \mathop {\lim }\limits_{x \to 3 + } \dfrac{{x + 3}}{{x - 3}} = \infty \]
\[\therefore \]The value of the function \[\dfrac{{x + 3}}{{x - 3}}\]as x approaches \[3 + \]is \[\infty \].

Note: Many students make the mistake of writing the limit value of \[\dfrac{6}{h} = 0\]and end up with a final answer as 1. Keep in mind any fraction with denominator as 0 is equal to infinity, the value of fraction becomes zero if numerator is 0. Many students get mistaken by the plus sign in the limit and think that we move in the positive direction on the number line and thus take values in direction from left to right. Students should focus on the fact that the limit of the function at a point can be reached from two sides, either from left or from right and when we move from right to the limit then it is called right hand limit.