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# check continuity and differentiability f(x)= {1/1+ e^1/x} , x is not equal to 0          0   , x=0

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
6 years ago
Hello Student,

Left hand limit(LHL) = Right hand limit(RHL) = f(0)
$\\LHL = \lim_{h\rightarrow 0}\frac{1}{1+e^{\frac{-1}{h}}} \\LHL = \lim_{h\rightarrow 0}\frac{e^{\frac{1}{h}}}{1+e^{\frac{1}{h}}} \\LHL = \lim_{h\rightarrow 0}\frac{e^{\frac{1}{h}}+1-1}{1+e^{\frac{1}{h}}}\\ LHL = \lim_{h\rightarrow 0}1-\frac{1}{1+e^{\frac{1}{h}}} = 1\\ RHL = \lim_{h\rightarrow 0}\frac{1}{1+e^{\frac{1}{h}}} = 0\\$
f(0) = 0
So, f(x) is not continuous at x = 0.
Similarly, you can check the left hand & right hand derivative.
$\lim_{h\rightarrow 0}\frac{f(-h)}{-h} = \lim_{h\rightarrow 0}\frac{f(h)}{h}$
Thanks & Regards
Arun Kumar
Btech, IIT Delhi