Samacheer Kalvi Class 11 Business Maths Solutions Chapter 5 Differential Calculus Exercise 5.3

Question 1. Examine the following functions for continuity at indicated points.
(a) \( f(x)= \begin{cases} \frac{x^2-4}{x-2} & \text{if } x \neq 2 \\ 0 & \text{if } x=2 \end{cases} \) at \( x=2 \)
(b) \( f(x) = \begin{cases} \frac{x^2-9}{x-3} & \text{if } x \neq 3 \\ 6 & \text{if } x=3 \end{cases} \) at \( x=3 \)
Answer:
(a) To check continuity at \( x=2 \), we need to find the limit of \( f(x) \) as \( x \) approaches 2 from the left side.
\( L[f(x)]_{x=2} = \lim_{x \to 2^{-}} f(x) \)
We substitute \( x = 2 - h \), where \( h \to 0 \).
\( = \lim_{h \to 0} f(2-h) \)
\( = \lim_{h \to 0} \frac{(2-h)^2 - 4}{(2-h) - 2} \)
\( = \lim_{h \to 0} \frac{4 - 4h + h^2 - 4}{-h} \)
\( = \lim_{h \to 0} \frac{h^2 - 4h}{-h} \)
\( = \lim_{h \to 0} \frac{h(h-4)}{-h} \)
\( = \lim_{h \to 0} -(h-4) \)
\( = \lim_{h \to 0} (4-h) \)
\( = 4 - 0 = 4 \)
The value of the function at \( x=2 \) is given as \( f(2) = 0 \).
Since the left-hand limit \( L[f(x)]_{x=2} = 4 \) is not equal to \( f(2) = 0 \), the function \( f(x) \) is not continuous at \( x=2 \). A function must have its limit equal to its value at the point to be continuous.
In simple words: For part (a), we looked at the function around \( x=2 \). The value the function gets very close to (its limit) was 4, but the actual value at \( x=2 \) was 0. Since these are different, the function has a break at \( x=2 \).
(b) To check continuity at \( x=3 \), we need to find both the left-hand limit and the right-hand limit of \( f(x) \) as \( x \) approaches 3.
Left-hand limit: \( L[f(x)]_{x=3} = \lim_{x \to 3^{-}} f(x) \)
Substitute \( x = 3 - h \), where \( h \to 0 \).
\( = \lim_{h \to 0} f(3-h) \)
\( = \lim_{h \to 0} \frac{(3-h)^2 - 9}{(3-h) - 3} \)
\( = \lim_{h \to 0} \frac{9 - 6h + h^2 - 9}{-h} \)
\( = \lim_{h \to 0} \frac{h^2 - 6h}{-h} \)
\( = \lim_{h \to 0} \frac{h(h-6)}{-h} \)
\( = \lim_{h \to 0} -(h-6) \)
\( = \lim_{h \to 0} (6-h) \)
\( = 6 - 0 = 6 \)
Right-hand limit: \( R[f(x)]_{x=3} = \lim_{x \to 3^{+}} f(x) \)
Substitute \( x = 3 + h \), where \( h \to 0 \).
\( = \lim_{h \to 0} f(3+h) \)
\( = \lim_{h \to 0} \frac{(3+h)^2 - 9}{(3+h) - 3} \)
\( = \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} \)
\( = \lim_{h \to 0} \frac{h^2 + 6h}{h} \)
\( = \lim_{h \to 0} \frac{h(h+6)}{h} \)
\( = \lim_{h \to 0} (h+6) \)
\( = 0 + 6 = 6 \)
The value of the function at \( x=3 \) is given as \( f(3) = 6 \).
Since the left-hand limit, the right-hand limit, and the function's value all equal 6, the function \( f(x) \) is continuous at \( x=3 \). This means there are no breaks or jumps in the function's graph at this point.
In simple words: For part (b), the function approaches the same value (6) from both the left and the right sides of \( x=3 \), and the actual value at \( x=3 \) is also 6. Since all three match, the function is smooth and continuous at \( x=3 \).

🎯 Exam Tip: To prove continuity at a point, always check three things: the left-hand limit, the right-hand limit, and the function's value at that point. All three must be equal.

Question 2. Show that \( f(x) = |x| \) is continuous at \( x = 0 \).
Answer: We know that the absolute value function \( f(x) = |x| \) can be written as:
\( f(x) = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \)
To show continuity at \( x=0 \), we need to check the left-hand limit, the right-hand limit, and the value of the function at \( x=0 \).
Left-hand limit: \( \lim_{x \to 0^{-}} f(x) \)
We substitute \( x = 0 - h \), where \( h \to 0 \).
\( = \lim_{h \to 0} f(0-h) = \lim_{h \to 0} f(-h) \)
Since \( -h < 0 \), we use \( f(x) = -x \).
\( = \lim_{h \to 0} -(-h) = \lim_{h \to 0} h = 0 \)
Right-hand limit: \( \lim_{x \to 0^{+}} f(x) \)
We substitute \( x = 0 + h \), where \( h \to 0 \).
\( = \lim_{h \to 0} f(0+h) = \lim_{h \to 0} f(h) \)
Since \( h > 0 \), we use \( f(x) = x \).
\( = \lim_{h \to 0} h = 0 \)
Value of the function at \( x=0 \):
\( f(0) = |0| = 0 \)
Since \( \lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0) = 0 \), the function \( f(x) = |x| \) is continuous at \( x=0 \). The absolute value function is known for its smooth transition at the origin, despite the change in its definition.
In simple words: The absolute value function \( |x| \) approaches 0 from both the left and right sides of \( x=0 \). Also, its actual value at \( x=0 \) is 0. Since all these match, the function is continuous (has no breaks) at \( x=0 \).

🎯 Exam Tip: When dealing with piecewise functions like \( |x| \), remember to use the correct definition for the left-hand and right-hand limits based on the inequality given for each piece.