Question

The function $ f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} $ is not defined at $ x = 3 $ . What should be assigned to \[f\left( 3 \right)\] for continuity of $ f(x) $ at $ x = 3 $ ?

Answer 1

Hint: Continuity at a point can be determined by substituting the point in the function. If this gives a definite value then the function is said to be continuous at that point and if it is not defined then it is not continuous.
At $ x = 3 $ the function, $ f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} $ is not defined, still we can find the limit at which function is continuous.
Find $ f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}} $ by factorization of the numerator of $ f(x) $ .
Use the factorization formula, $ {a^2} - {b^2} = (a - b)(a + b) $ .
Next, substitute $ x = 3 $ for continuity of $ f(x) $ .

Complete step-by-step answer:
Consider the function, $ f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} $ .
It is given that the function is not continuous at $ x = 3 $ as the denominator becomes zero when we put $ x = 3 $ .
We have to find the value \[f\left( 3 \right)\] for continuity of $ f(x) = \dfrac{{{x^2} - 9}}{{x - 3}} $ .
To be the function $ f(x) $ continuous, find the limit,
 $ f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - 9}}{{x - 3}} $
Write $ 9 $ as $ {3^2} $ .
 $ f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{{x^2} - {3^2}}}{{x - 3}} \ldots \ldots (1) $
Apply the factorization formula, $ {a^2} - {b^2} = (a - b)(a + b) $ to the numerator of the right-hand side.
Substitute $ a = x $ and $ b = 3 $ into the factorization formula and $ {a^2} - {b^2} = (a - b)(a + b) $ ,
 $ {x^2} - {3^2} = (x - 3)(x + 3) \ldots \ldots (2) $
From equation $ (1) $ and $ (2) $ .
 $ f(3) = \mathop {\lim }\limits_{x \to 3} \dfrac{{(x - 3)(x + 3)}}{{x - 3}} $
Cancel out the $ x - 3 $ numerator of the right-hand side.
 $ f(3) = \mathop {\lim }\limits_{x \to 3} x + 3 $
Apply the limit as $ x $ tends to $ 3 $ ,
 $ f(3) = 3 + 3 $
 $ f(3) = 6 $

Final Answer: The value assigned to \[f\left( 3 \right)\] for continuity of $ f(x) $ at $ x = 3 $ is $ 6 $ .

Note:
It is important to remember the definition of continuity
A function $ f(x) $ is said to be continuous at a point \[x = a\] if $ \mathop {\lim }\limits_{x \to a} f(x) = f(a) $ and
If $ f(x) $ is continuous at \[x = a\] then,
Limit at the point: $ \mathop {\lim }\limits_{x \to a} f(x) = f(a) $
Left-hand limit: $ \mathop {\lim }\limits_{x \to {a^ - }} f(x) = f(a) $
Right-hand limit: $ \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a) $