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	If `f(x)` = { `x/(1 + e^(1/x))` for `x != 0`, 0 for x = 0, then the function f(x) is different... (Doubtnut)  View
	If f(x) = {{:(x((e^(1//x) - e^(-1//x))/(e^(1//x)+e^(1//x))),,x ne 0),(0,,x = 0):}, then at x = 0... (Doubtnut)  View
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