Assuming is differentiable at 0, the limit
converges and is equal to ; which yields .
Here we go then!
You have . Hence either or is identically zero which instantly yields . Hence suppose .
is easy; set x=y.
is also easy.
by the above.
Now consider the complex numbers
.
We have .
In particular , and .
Hence the function is a morphism of groups . Hence we have for some . Clearly since . Hence
Hence ,
so that .
Just a little extra :
In case you are wondering why expressing m as an exponential map is possible, consider
Then for all integers n:
so that
from this it follows easily that for all rational x
.
Hence if is continuous on any interval, it coincides everywhere on that interval with (irrationals included).
But is certainly continuous at 0 (it's differentiable there!). To see that g is continuous at 0, we have
i.e.
and from this it is easy to show (with sequences for example) that is continuous at 0 (using the fact that is continuous at 0 and is equal to 0)