Rajasthan Board RBSE Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.1
Question 1.
Show that left and right limits of function
at x = 1 are equal and their value is 1.
Solution:
From equation (i) and (ii)
L.H.L. = R.H.L.
⇒ f(1 – 0) = f(1 + 0) = 1
Hence, left and right limits of the given function at x = 1 are equal and their value is 1.
Hence Proved.
Question 2.
Is limit of the function at x = 0 ?
Solution:
It is clear from equation (i) and (ii),
L.H.L. ≠ R.H.L.
⇒ f(0 – 0) ≠ x(0 + 0)
Hence, limit of function does not exist at x = 0.
Question 3.
Prove that at x = 0, limits of function f(x) = | x | + | x – 1| exists.
Solution:
f(x) = | x | + | x – 1 |
Right limit at x – 0
R.H.L. = limx→0+ + f(x) = f(0 + 0)
= limh→0 f(0 + h)
= limh→0 |0 + h| + |0 + h – 1 | (h> 0)
= limh→0 | h | + | h – 1 |
= 0 + | 0 – 1 | = 1 …(i)
Left limit at x = 0
L.H.L. = limh→0– + f(x) = f(0 – 0)
= limh→0 f(0 – h)
= limh→0 | 0 – h | + | 0 – h – 1 |
= limh→0 | 0 – h | + | -(h – 1) |
= | -0 | + | -(0+ 1) |
= 0 + 1 = 1 …(ii)
It is clear from equation (i) and (ii),
L.H.L. = R.H.L.
⇒ f(0 – 0) = f(0 + 0) = 1
Hence, limit of function exists at x = 0. Hence Proved.
Question 4.
Prove that at x = 2, limits of function does not exists.
Solution:
Right limit at x = 2
R.H.L. = limx→0 + f(x)
= f(2 + 0) = limh→0 f(2 + h)
= limh→0[(2 + h)2 + (2 + h) + 1]
= limh→0 [4 + h2 + 4h + 2 + h + 1]
= limh→0 [h2 + 5h + 7]
= 02 + 5(0) + 7 = 7 ….(i)
Left limit at x = 2
L.H.L. = limx→0 – f(x)
= f(2 – 0) = limh→0 f(2 – h)
= limh→0 [2 – h]
= 2 – 0 = 2 .
From equation (i) and (ii),
L.H.L. ≠ R.H.L.
⇒ f(2 – 0) ≠ f(2 + 0)
Hence, limit of function does not exist at x = 2.
Question 5.
Find the left and right limit of function f(x) = x cos ( 1/x) at x = 0.
Solution: