RBSE Solutions for Class 11 Maths Chapter 10 Limits and Derivatives Ex 10.1

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. = lim_x→0⁺ + f(x) = f(0 + 0)
= lim_h→0 f(0 + h)
= lim_h→0 |0 + h| + |0 + h – 1 | (h> 0)
= lim_h→0 | h | + | h – 1 |
= 0 + | 0 – 1 | = 1 …(i)
Left limit at x = 0
L.H.L. = lim_h→0^– + f(x) = f(0 – 0)
= lim_h→0 f(0 – h)
= lim_h→0 | 0 – h | + | 0 – h – 1 |
= lim_h→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. = lim_x→0 + f(x)
= f(2 + 0) = lim_h→0 f(2 + h)
= lim_h→0[(2 + h)² + (2 + h) + 1]
= lim_h→0 [4 + h² + 4h + 2 + h + 1]
= lim_h→0 [h² + 5h + 7]
= 0² + 5(0) + 7 = 7 ….(i)
Left limit at x = 2
L.H.L. = lim_x→0 ^– f(x)
= f(2 – 0) = lim_h→0 f(2 – h)
= lim_h→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:

RBSE Solutions for Class 11 Maths