Is it true that for complex $s$,
$\lim_{s\to 1} \frac{2-2^s}{s-1}=-2\log (2)?$
If so, prove it.
Is it true that for complex $s$,
$\lim_{s\to 1} \frac{2-2^s}{s-1}=-2\log (2)?$
If so, prove it.
Yes it is true.
Added: The nicest solution is to just note that the above is the definition of the derivative.
Hint: Use l'Hopitals rule. Make sure you can justify why this is allowed now that we are dealing with complex numbers.
Alternative Hint: Use power series. Since $2^s$ is analytic everywhere, we can expand around $s=1$ and write it as $2e^{(s-1)\log 2}=2+2(s-1)\log 2+2\frac{(s-1)^2}{2!}\log^2 2+\cdots $ From this you can deduce the Laurent series around $s=1$ for $\frac{2-2^s}{s-1}.$