Question

f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0 

Answer 1

We have, f(x) = `{{:(("e"^(1/x))/(1 + "e"^(1/x))",", "if"  x ≠ 0),(0",", "if"  x = 0):}` at x = 0 

At x = 0

L.H.L. = `lim_(x -> 0^-) ("e"^(1/x))/(1 + "e"^(1/x))`

= `lim_("h" -> 0) ("e"^(1/(0 - "h")))/(1 + "e"^(1/(0 - "h"))`

= `lim_("h" -> 0) ("e"^(1/"h"))/(1 + "e"^(- 1/"h"))`

= `("e"^(- oo))/(1 + "e"^(- oo))`

= `0/(1 + 0)`

= 0

R.H.L. = `lim_(x -> 0^+) ("e"^(1/x))/(1 + "e"^(1/x))`

= `lim_("h" -> 0) ("e"^(1/(0 + "h")))/(1 + "e"^(1/(0 + "h"))`

= `lim_("h" -> 0) ("e"^(1/"h"))/(1 + "e"^(1/"h"))`

= `lim_("h" -> 0) 1/("e"^(-1/"h") + 1)`

= `1/("e"^(-oo) + 1)`

= `1/(0 + 1)`

= 1

Thus, L.H.L. ≠ R.H.L. at x = 0.

So, f(x) is discontinuous at x = 0.