Evaluate $\lim_{x\to\alpha^+}\frac{2\ln(\sqrt x-\sqrt{\alpha})}{\ln(e^{\sqrt x}-e^{\sqrt{\alpha}})}$ without L'Hopital
Using L'Hopital, I am getting $2$ as answer.
I wonder how to do it without L'Hopital.
Attempt $1$:
In numerator, taking $2$ inside, I get, $\ln(x+\alpha-2\sqrt{x\alpha})=\ln(\alpha(1+\frac x{\alpha}-2\sqrt{\frac x{\alpha}}))$
Not sure what to do with it.
Attempt $2$:
Writing denominator as $\ln(e^{\sqrt{\alpha}}(e^{\sqrt x-\sqrt{\alpha}}-1))=\sqrt{\alpha}+\ln(\sqrt x-\sqrt{\alpha})$
Not able to proceed next.
Attempt $3$:
Putting $x=\alpha+h$,
Numerator=$2\ln(\sqrt{\alpha+h}-\sqrt{\alpha})=2\ln(\sqrt{\alpha}(\sqrt{1+\frac h{\alpha}}-1))$
Writing $\sqrt{1+\frac h{\alpha}}$ as $1+\frac h{2\alpha}$, but this isn't helping either.
