# Q.Can we say that all even functions are continuous and non differentiable at x=0 and all odd functions are continuous and differentiable at x=0:?

Arun Kumar IIT Delhi
9 years ago
Hello Student,

No we can’t generally say that.
In even function take the example of cosx
its continous and differentiable at 0.
In odd function take example of sin(1/x).
its non continous and non differentiable at 0.

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Jitender Singh IIT Delhi
9 years ago
Ans: Yes
Sol:
Let f(x) be the even function, We have
$f(x)= f(-x)$
$f(0)= 0$
Proof for continuity:
Right Hand Limit (RHL) = f(0) = Left Hand Limit (LHL)
$\lim_{h\rightarrow 0}f(h) = 0 = \lim_{h\rightarrow 0}f(-h)$
$RHL = \lim_{h\rightarrow 0}f(h) = f(0) = 0$
$LHL = \lim_{h\rightarrow 0}f(-h) = f(0) = 0$
Proof for Non Differentiability:
For f to be differentiable,
Left Hand Derivative (LHD) = Right Hand Derivative (RHD)
$\lim_{h\rightarrow 0}\frac{f(-h)-f(0)}{-h} = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h}$
$LHD = \lim_{h\rightarrow 0}\frac{f(-h)-f(0)}{-h} = \lim_{h\rightarrow 0}-\frac{f(h)}{h}$
$RHD = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} = \lim_{h\rightarrow 0}\frac{f(h)}{h}$
$LHD \neq RHD$
Let f(x) be odd,
$f(x)= -f(-x)$
$f(0)= 0$
Proof for continuity:
RHL = f(0) = LHL
$RHL = \lim_{h\rightarrow 0}f(h) = f(0) = 0$
$LHL = \lim_{h\rightarrow 0}f(-h) = f(0) = 0$
Proof for Differentiability:
LHD = RHD
$LHD = \lim_{h\rightarrow 0}\frac{f(-h)-f(0)}{-h} = \frac{-f(h)}{-h} = \frac{f(h)}{h}$
$RHD = \lim_{h\rightarrow 0}\frac{f(h)-f(0)}{h} = \frac{f(h)}h}$
Thanks & Regards
Jitender Singh
IIT Delhi
Jitender Singh
13 Points
9 years ago
There is a little mistake in the previous solution.
In case of even, differentiability depend on the f(x). For example,
f(x) = x2,  LHD = RHD = 0
f(x) = |x|, LHD = -1   RHD = 1
x2 is differentiable but |x| is not.
Similarly for the odd function. For example,
f(x) = x3, LHD = 0 RHD = 0  Differentiable
f(x) = x1/3, Derivative becomes infinite. So non differentiable

Thanks & Regards
Jitender Singh
IIT Delhi