Question

F(x)=(e^1/x +1)/(e^1/x -1) is function continuous at x=0?

Do you mean `F(x)=(e^(1/x) +1)/(e^(1/x) -1)`

Answer 1

No. Lateral limits are different at `x_0=0`

Explanation:

`lim_(xrarr0^-)f(x)` `=` `lim_(xrarr0^-)(e^(1/x)+1)/(e^(1/x)-1)` `=-1`

`lim_(xrarr0^+)f(x)` `=` `lim_(xrarr0^+)(e^(1/x)+1)/(e^(1/x)-1)` `= 1`

so `lim_(xrarr0^-)f(x)` `!=` `lim_(xrarr0^+)f(x)` and `f` is as a result not continuous at `x_0=0`

• because

`lim_(xrarr0^-)(e^(1/x)+1)/(e^(1/x)-1)`

`1/x=y`
`x->0^-`
`y->-oo`

`=` `lim_(yrarr-oo)(e^y+1)/(e^y-1)` `= (0+1)/(0-1) = -1`

`lim_(xrarr0^+)(e^(1/x)+1)/(e^(1/x)-1)`

`1/x=u`
`x->0+`
`u->+oo`

`=` `lim_(urarr+oo)(e^u+1)/(e^u-1)` -- `((+oo)/(+oo))`

and with rules De L'Hopital

`=` `lim_(urarr+oo)(e^u)/(e^u)` `=` `lim_(urarr+oo)1` `=1`

(Note: you can check in the graph the behaviour of `e^x` while `x->+-oo` )

graph{e^x [-22.79, 22.82, -11.42, 11.37]}

Answer 2

`f(x) = e^(1/(x+1))/e^(1/(x-1))` is continuous at `0`.

Explanation:

Both `1/(x+1)` and `1/(x-1)` are continuous at `0`, so

both `e^(1/(x+1))` and `e^(1/(x-1))` are continuous at `0`

Because `e^(1/(x-1)) != 0` at `x=0`,

the quotient `e^(1/(x+1))/e^(1/(x-1))` is also continuous at `0`.