Examine the differentiability of f, where f is defined by

Given,
…(1)
We need to check whether f(x) is continuous and differentiable at x = 0
A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-
![]()
Where h is a very small number very close to 0 (h→0)
And a function is said to be differentiable at x = c if it is continuous there and
Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).
Mathematically we can represent it as-
![]()
![]()
Finally, we can state that for a function to be differentiable at x = c
![]()
Checking for the continuity:
Now according to above theory-
f(x) is continuous at x = 0 if -
![]()
∴ LHL = ![]()
⇒ LHL =
{using equation 1}
As sin (-1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHL = 02 × (finite value) = 0
∴ LHL = 0 …(2)
Similarly,
RHL = ![]()
⇒ RHL =
{using equation 1}
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHL = (0)2(finite value) = 0 …(3)
And, f(0) = 0 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:
![]()
∴ f(x) is continuous at x = 0. So we will proceed now to check the differentiability.
Checking for the differentiability:
Now according to above theory-
f(x) is differentiable at x = 0 if -
![]()
∴ LHD = ![]()
⇒ LHD =
{using equation 1}
⇒ LHD = ![]()
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ LHD = 0×(some finite value) = 0
∴ LHD = 0 …(5)
Now,
RHD = ![]()
⇒ RHD =
{using equation 1}
⇒ RHD = ![]()
As sin (1/h) is going to be some finite value from -1 to 1 as h→0
∴ RHD = 0×(some finite value) = 0
∴ RHD = 0 …(6)
Clearly from equation 5 and 6,we can conclude that-
(LHD at x=0) = (RHD at x = 0)
∴ f(x) is differentiable at x = 0