Question

How to test continuty of this function ? f(x) = `(1+x)^(1+x)` x is not equal zero

Answer 1

`f(x)` is continuous for every `x in (-1,+oo)`.

Explanation:

Write the function as:

`f(x) = e^((1+x)ln(1+x))`

As the exponential function `e^t` is continuous for any `t in RR`, and `(1+x)ln(1+x)` is continuous for `x > -1` then `f(x)` is continuous for every `x in (-1,+oo)` and `x!=0`

For `x=0`:

`lim_(x->0) f(x) = lim_(x->0) e^((1+x)ln(1+x)) = e^((lim_(x->0)(1+x)ln(1+x))) = e^0 =1 = f(0)`

So the function is continuous also for `x = 0`

graph{(1+x)^(1+x) [-10, 10, -5, 5]}

Answer 2

This function is not continuous at `x=0`

Explanation:

Consider a function `g(h)=(1+1/h)^h`.

This function approrches `e(=2.718…)` when `h`-> `oo` (by definition of `e`). This is also true when `h` goes to negative infinity.

So, the limit of `f(x)` when `x` approaches 0 is
`lim_(x->+0) f(x)` = `lim_(h->oo) g(h)` =`e`
`lim_(x->-0) f(x)` = `lim_(h->-oo) g(h)` =`e`

They are different from `f(0)` and therefore `f(x)` has a jump
discontinuity at `x=0`.