Question

Which values of x this function is continuous? (x non-negative real number and n->infinity)

`f(x)=limlog(e^n+x^n)/n`

Answer 1

For `x in (0,e)`, write the function under the limit as:

`log (e^n+x^n)/n = log (e^n(1+(x/e)^n))/n`

and using the properties of logarithms:

`log (e^n+x^n)/n = (log (e^n)+ log(1+(x/e)^n))/n`

`log (e^n+x^n)/n =1+ ( log(1+(x/e)^n))/n`

Now, as `x/e < 1` we have `lim_(n->oo) (x/e)^n = 0`, so `f(x) = 1`

For `x = e`, we have that:

`log (e^n+x^n)/n = log (2e^n)/n = (log2+n)/n`

and again `f(x) = 1`

For `x > e` write the function as:

`log (e^n+x^n)/n = log (x^n((e/x)^n+1))/n`

`log (e^n+x^n)/n = (nlogx+log((e/x)^n+1))/n`

`log (e^n+x^n)/n = logx+log((e/x)^n+1)/n`

We have now `e/x < 1` and `lim_(n->oo) (e/x)^n =0`, so `f(x) = logx`

We can conclude that the function `f(x)` is:

`f(x) = {(1 " for " x <=e),(logx " for " x > e):}`

and is therefore continuous for every `x in RR`

Answer 2

It is continuous for all non-negative real `x`

Explanation:

It looks like the function is:

`f(x) = lim_(nrarroo)ln(e^n+x^n)/n`

In the following `x` always represents a non-negative real number.

For every `x` `lim_(nrarroo)(e^n+x^n) = oo` so the defining limit above has indeterminate form `oo/oo`.

Apply l"Hospital's Rule (differentiate with respect to `n`) and find

`lim_(nrarroo)(e^n+x^nlnx)/(e^n+x^n)`

Case 1 For `x < e` so `x/e < 1` divide numerator and denominator by `e` to get

`(1+(x/e)^nlnx)/(1+(x/e)^n` whose limit at `oo` is `1`.

Case 2 For `x=e`, we have `lim_(nrarroo) (e^n+e^n)/(e^n+e^n) = lim_(nrarroo)1 = 1`

Case 3 For `x > e` so `e/x < 1` divide numerator and denominator by `e` to get

`((e/x)^n+lnx)/((e/x)^n+1)` whose limit at `oo` is `lnx`.

So,

`f(x) = {(1,"if",x<=e),(lnx,"if",x>e):}` which is continuous at all `x >=0`