problematic limit with squares I have problem with calculation such limit : $\displaystyle \lim_{n\to \infty} {\frac{10^{\sqrt{n+1}-\sqrt{n}}-1}{2^{\frac{1}{\sqrt{n}}}-1}}$ 
I only know the answer from wolfram that it's $\frac{1}{2}+\frac{\ln{5}}{\ln{4}}$

First of all
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.$$
Now, write $x=1/\sqrt{n}.$ Thus, $\sqrt{n+1}=\frac{\sqrt{x^2+1}}{x}.$
Now, we compute the limit
$$\displaystyle \lim_{n\to \infty} {\frac{10^{\sqrt{n+1}-\sqrt{n}}-1}{2^{\frac{1}{\sqrt{n}}}-1}}=\lim_{x\to 0^+} \frac{10^{\frac{x}{1+\sqrt{x^2+1}}}-1}{2^x-1} \underbrace{=}_{\mathrm{L'Hospital}} \lim_{x\to 0^+} \frac{10^{\frac{x}{1+\sqrt{x^2+1}}} \frac{1}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} \ln 10}{2^x\ln 2}\\=\frac{\frac12 \ln 10}{\ln 2}\underbrace{=}_{\ln 10=\ln (2\cdot 5)=\ln 2+\ln 5}\frac{\frac12 \ln 2+\frac 12 \ln 5}{\ln 2}=\frac12+\frac{\ln 5}{2\ln 2}=\frac12+\frac{\ln 5}{\ln 2^2}=\frac12+\frac{\ln 5}{\ln 4}.$$
