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# How to show a function is continuous on R+ watch

1. Hello to everyone
I have tried to make a short exercise, but it appears difficult in the end.
I put the exercise and what i have made here.
If someone can help me, i will be so thankful.
Best regard to everyone

Exercise: Tell if the function x →x^(1+1/x) is defined in (R*)+, study its variations, its limits
to the terminals and to say if it admits a continuous extension on R +. Finally, say if it admits a
Extension on R +.

- We know that 1/x is defined on R*

-We know that x is defined on R so x is also defined on R+.

Then x^(1+1/x) is defined on (R*)+

-We know 1/x is decreasing on ]0;+∞[

So 1+ 1/x is decreasing

-We know x is increasing on ]0 ; +∞[

Then the global function x x^(1+1/x) is decreasing on (R*)+

-lim 1+1/x= 1

x→+∞

-lim x=1

x→1

Then lim x^(1+1/x) =1

x→+∞

-lim 1+ 1/x= +∞

x→0

-lim x=+∞

x→+∞

Then lim x^(1+1/x) =+∞

x→+∞

let's see if f admits a continuous extension on R+

f(x) is extensible by continuity in 0 if f(x) has a right or left limit next to 0.
We extend f by continuity by creating a function f' equal to f on R + such that f'(0)=0

Yeah that's the problem I don't know how to continue
2. (Original post by PotterGranger)
Hello to everyone
I have tried to make a short exercise, but it appears difficult in the end.
I put the exercise and what i have made here.
If someone can help me, i will be so thankful.
Best regard to everyone

Exercise: Tell if the function x →x^(1+1/x) is defined in (R*)+, study its variations, its limits
to the terminals and to say if it admits a continuous extension on R +. Finally, say if it admits a
Extension on R +.

- We know that 1/x is defined on R*

-We know that x is defined on R so x is also defined on R+.

Then x^(1+1/x) is defined on (R*)+

-We know 1/x is decreasing on ]0;+∞[

So 1+ 1/x is decreasing

-We know x is increasing on ]0 ; +∞[

Then the global function x x^(1+1/x) is decreasing on (R*)+

-lim 1+1/x= 1

x→+∞

-lim x=1

x→1

Then lim x^(1+1/x) =1

x→+∞

-lim 1+ 1/x= +∞

x→0

-lim x=+∞

x→+∞

Then lim x^(1+1/x) =+∞

x→+∞

let's see if f admits a continuous extension on R+

f(x) is extensible by continuity in 0 if f(x) has a right or left limit next to 0.
We extend f by continuity by creating a function f' equal to f on R + such that f'(0)=0

Yeah that's the problem I don't know how to continue
Since the limit as x approaches 0 of f(x) is not defined, there is no continuous extension.
3. (Original post by HapaxOromenon3)
Since the limit as x approaches 0 of f(x) is not defined, there is no continuous extension.
Isn't the limit at 0 from the right equal to 0?
4. (Original post by RichE)
Isn't the limit at 0 from the right equal to 0?
If the limit exists from both sides then you define the continuous extension piecewise as equal to the original function for all values of x other than the discontinuity, and equal to the numerical value of the limit when x is the discontinuous point.
5. (Original post by HapaxOromenon3)
If the limit exists from both sides then you define the continuous extension piecewise as equal to the original function for all values of x other than the discontinuity, and equal to the numerical value of the limit when x is the discontinuous point.
Not sure what your point is. In the above example we have a function defined for positive x.

It has a limit of 0 at 0 so can be continuously extended to nonnegative x.

It can be extended continuously to the real line by making it an even function.

PS that said I may have misunderstood the OP's set notation which is pretty nonstandard.
6. (Original post by RichE)
Not sure what your point is. In the above example we have a function defined for positive x.

It has a limit of 0 at 0 so can be continuously extended to nonnegative x.

It can be extended continuously to the real line by making it an even function.
Or, indeed just by making it 0 for x<=0.

PS that said I may have misunderstood the OP's set notation which is pretty nonstandard.

I hate to say it, but I'm also a little skeptical that this is a question the OP "tried to make up as a short exercise", which makes me reticent to give too much help.

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