# Image is attached . The function fx is discontinous at ?

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
$f(x) = [x^{2}]-[x]^{2}$
Lets check at x = 0
$LHL =\lim_{h\rightarrow 0} ([(-h)^{2}]-[-h]^{2}) = -1$
$RHL =\lim_{h\rightarrow 0} ([(h)^{2}]-[h]^{2}) = 0$
$f(0) = 0$
So, f(x) is discontinuous at x = 0
Lets check at x = 1
$LHL =\lim_{h\rightarrow 0} ([(1-h)^{2}]-[1-h]^{2}) = 0$
$RHL =\lim_{h\rightarrow 0} ([(1+h)^{2}]-[1+h]^{2}) = 0$
$f(1) = 0$
So, f(x) is continuous at x = 1
Lets check at x = 2
$LHL =\lim_{h\rightarrow 0} ([(2-h)^{2}]-[2-h]^{2}) = 3-1 = 2$
$RHL =\lim_{h\rightarrow 0} ([(2+h)^{2}]-[2+h]^{2}) = 4-4 = 0$
$f(2) = 0$
You should have given the options. Otherwise, you can check like this.
Thanks & Regards
Jitender Singh
IIT Delhi