StudentAidUnit 12: Limit and Continuity — Exercise 12 2
11th Class Mathematics · Unit 12: Limit and Continuity
1.(i) Determine the left hand limit and the right hand limit and then find the limit of f(x) = 2x² + x - 5 when x → 1 (ii) Determine the left hand limit and the right hand limit and then find the limit of f(x) = (x² - 9)/(x - 3) when x → -3 (iii) Determine the left hand limit and the right hand limit and then find the limit of f(x) = |x - 5| when x → 5
Given
(i) Function
f(x) = 2x2 + x - 5
Formula
Left-hand limit
Limx to 1^- f(x) = Limx to 1^-(2x2 + x - 5)
Working
= 2(1)2 + 1 - 5 = 2 + 1 - 5
Result
LHL
= -2
Formula
Right-hand limit
Limx to 1^+ f(x) = Limx to 1^+(2x2 + x - 5)
Working
= 2(1)2 + 1 - 5 = 2 + 1 - 5
Result
RHL
= -2
As Limx to 1^- f(x) = Limx to 1^+ f(x)
Limit
Limx to 1 f(x) = -2
Given
(ii) Function
f(x) = x2 - 9x - 3, quad c = -3
Formula
Left-hand limit
Limx to -3^- f(x) = Limx to -3^- x2 - 9x - 3
Working
= (-3)2 - 9-3 - 3 = 9 - 9-6 = 0-6
Result
LHL
= 0
Formula
Right-hand limit
Limx to -3^+ f(x) = Limx to -3^+ x2 - 9x - 3
Working
= (-3)2 - 9-3 - 3 = 9 - 9-6 = 0-6
Result
RHL
= 0
As Limx to -3^- f(x) = Limx to -3^+ f(x)
Limit
Limx to -3 f(x) = 0
Given
(iii) Function
f(x) = |x - 5|, quad c = 5
Formula
Absolute value expansion
|x - 5| = begin{cases} +(x - 5) & if x geq 5 -(x - 5) & if x < 5 end{cases}
Left-hand limit
Limx to 5^- f(x) = Limx to 5^- |x - 5|
Working
= Limx to 5^- [-(x - 5)] = -(5 - 5)
Result
LHL
= 0
Formula
Right-hand limit
Limx to 5^+ f(x) = Limx to 5^+ |x - 5|
Working
= Limx to 5^+ (x - 5) = 5 - 5
Result
RHL
= 0
As Limx to 5^- f(x) = Limx to 5^+ f(x)
Limit
Limx to 5 |x - 5| = 0
2.(i) Discuss the continuity of f(x) = {2x + 5 if x ≤ 2; 4x + 1 if x > 2} at x = 2 (ii) Discuss the continuity of f(x) = {3x - 1 if x < 1; 4 if x = 1; 2x if x > 1} at x = 1
Given
(i) At x = 2
f(x) = 2x + 5 ⇒ f(2) = 2(2) + 5
Result
Function value
f(2) = 9
Formula
Left-hand limit
Limx to 2^- f(x) = Limx to 2^- (2x + 5)
Working
= 2(2) + 5
Result
LHL
= 9
Formula
Right-hand limit
Limx to 2^+ f(x) = Limx to 2^+ (4x + 1)
Working
= 4(2) + 1
Result
RHL
= 9
As Limx to 2^- f(x) = Limx to 2^+ f(x) = f(2) = 9
Conclusion
f(x) is continuous at x = 2
Given
(ii) At x = 1
f(x) = 4 ⇒ f(1) = 4
Result
Function value
f(1) = 4
Formula
Left-hand limit
Limx to 1^- f(x) = Limx to 1^- (3x - 1)
Working
= 3(1) - 1
Result
LHL
= 2
Formula
Right-hand limit
Limx to 1^+ f(x) = Limx to 1^+ (2x)
Working
= 2 Limx to 1^+ x
Result
RHL
= 2
As Limx to 1^- f(x) = Limx to 1^+ f(x)
so Limx to 1 f(x) exists
As f(1) = 4 neq 2 = Limx to 1 f(x)
Conclusion
f(x) is discontinuous at x = 1
3.(i) If f(x) = {3x if x ≤ -2; x² - 1 if -2 < x < 2; 3 if x ≥ 2}, discuss continuity at (i) x = 2 and (ii) x = -2
Given
(i) (i) At x = 2
f(x) = 3 ⇒ f(2) = 3
Formula
Left-hand limit
Limx to 2^- f(x) = Limx to 2^- (x2 - 1)
Working
= 22 - 1 = 4 - 1
Result
LHL
= 3
Formula
Right-hand limit
Limx to 2^+ f(x) = Limx to 2^+ (3)
Result
RHL
= 3
As Limx to 2^- f(x) = Limx to 2^+ f(x) = f(2) = 3
At x = 2
f(x) is continuous
Given
(ii) At x = -2
f(x) = 3x ⇒ f(-2) = 3(-2) = -6
Formula
Left-hand limit
Limx to -2^- f(x) = Limx to -2^- (3x)
Working
= 3(-2)
Result
LHL
= -6
Formula
Right-hand limit
Limx to -2^+ f(x) = Limx to -2^+ (x2 - 1)
Working
= (-2)2 - 1 = 4 - 1
Result
RHL
= 3
As Limx to -2^- f(x) = -6 neq 3 = Limx to -2^+ f(x)
At x = -2
f(x) is discontinuous
4.(i) If f(x) = {x + 2 if x ≤ -1; c + 2 if x > -1}, find c so that lim f(x) exists at x = -1
Formula
(i) For limit to exist
Limx to -1^- f(x) = Limx to -1^+ f(x)
Left-hand limit
Limx to -1^- (x + 2) = Limx to -1^- (x + 2)
Working
= -1 + 2
Result
LHL
= 1
Formula
Right-hand limit
Limx to -1^+ (c + 2) = Limx to -1^+ (c + 2)
Result
RHL
= c + 2
-1 + 2 = c + 2
1 = c + 2
Solution
c = -1
5.(i) Find the values m and n so that given function f(x) = {mx if x < 3; n if x = 3; -2x + 9 if x > 3} is continuous at x = 3 (ii) Find the value m so that given function f(x) = {mx if x < 3; x² if x ≥ 3} is continuous at x = 3
Given
(i) At x = 3
f(3) = n
Formula
Left-hand limit
Limx to 3^- f(x) = Limx to 3^- (mx)
Working
= m(3)
Result
LHL
= 3m
Formula
Right-hand limit
Limx to 3^+ f(x) = Limx to 3^+ (-2x + 9)
Working
= -2(3) + 9 = -6 + 9
Result
RHL
= 3
As f(x) is continuous at x = 3
So f(3) = Limx to 3^- f(x) = Limx to 3^+ f(x)
n = 3m = 3
Values
n = 3 and 3m = 3 ⇒ m = 1
Given
(ii) At x = 3
f(x) = x2 ⇒ f(3) = 32
Result
Function value
f(3) = 9
Formula
Left-hand limit
Limx to 3^- f(x) = Limx to 3^- (mx)
Working
= 3m
Formula
Right-hand limit
Limx to 3^+ f(x) = Limx to 3^+ (x2)
Working
= 32
Result
RHL
= 9
As f(x) is continuous at x = 3
So f(3) = Limx to 3^- f(x) = Limx to 3^+ f(x)
9 = 3m = 9
Solution
3m = 9 ⇒ m = 3
6.(i) If f(x) = {(√(2x + 5) - √(x + 7))/(x - 2) if x ≠ 2; k if x = 2}, find the value of k so that f is continuous at x = 2
Given
(i) At x = 2
f(x) = k ⇒ f(2) = k
Formula
Limit calculation
Limx to 2 f(x) = Limx to 2 frac{sqrt{2x + 5} - sqrt{x + 7}{x - 2}
Rationalize numerator
= Limx to 2 frac{sqrt{2x + 5} - sqrt{x + 7}{x - 2} × frac{sqrt{2x + 5} + sqrt{x + 7}{sqrt{2x + 5} + sqrt{x + 7}
= Limx to 2 frac{(2x + 5) - (x + 7)}{(x - 2)(sqrt{2x + 5} + sqrt{x + 7})}
= Limx to 2 frac{x - 2}{(x - 2)(sqrt{2x + 5} + sqrt{x + 7})}
= Limx to 2 frac{1}{sqrt{2x + 5} + sqrt{x + 7}
Working
= frac{1}{sqrt{2(2) + 5} + sqrt{2 + 7}
= frac{1}{sqrt{9} + sqrt{9}
= 13 + 3
Result
Limit value
Limx to 2 f(x) = 16
As f(x) is continuous at x = 2
Solution
k = 16
7.(i) Given the function f(x) = {2x + 3 if x ≤ 1; -x + 4 if x > 1}, discuss the limit and continuity at x = 1
Given
(i) For limit of f(x) at x = 1
First, find the left and right hand limits
Formula
Left-hand limit
Limx to 1^- f(x) = Limx to 1^- (2x + 3)
Working
= 2(1) + 3 = 2 + 3
Result
LHL
= 5
Formula
Right-hand limit
Limx to 1^+ f(x) = Limx to 1^+ (-x + 4)
Working
= -1 + 4
Result
RHL
= 3
Since Limx to 1^- f(x) neq Limx to 1^+ f(x)
Conclusion on limit
Limx to 1 f(x) does not exist
When Limx to 1 f(x) does not exist
Conclusion on continuity
f(x) is discontinuous at x = 1