Question

Check the function f(x) = lim_(x → 0) (e^1/x - 1)/(e^1/x + 1) for continuity and differentiability at x = 0.

Answer 1

[Dividing numerator and denominator by e^1/h]

= (1 - 0)/(1 + 0) = 1 [using Eq.(1)]

Clearly, lim_(x →0^-) f(x) lim_(x →0⁺)f(x)

Hence, lim_(x →0) f(x) does not exist.

As lim_(x →0) f(x) does not exist, function is neither continuous nor differentiable at x = 0.