<s>[INST] 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 $\frac{1}{2}+\frac{\ln{5}}{\ln{4}}$
 [/INST] We need to understand that both $$x = \sqrt{n + 1} - \sqrt{n} = \frac{1}{\sqrt{n + 1} + \sqrt{n}}$$ and $y = 1/\sqrt{n}$ tend to $0$ as $n \to \infty$ and we aslo know that $$\lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a$$ for $a > 0$. Hence we have $$\begin{aligned}l &= \lim_{n \to \infty}\frac{10^{\sqrt{n + 1} - \sqrt{n}} - 1}{2^{1/\sqrt{n}} - 1}\\
&= \lim_{n \to \infty}\frac{10^{\sqrt{n + 1} - \sqrt{n}} - 1}{\sqrt{n + 1} - \sqrt{n}}\cdot\frac{\sqrt{n + 1} - \sqrt{n}}{1/\sqrt{n}}\cdot\frac{1/\sqrt{n}}{2^{1/\sqrt{n}} - 1}\\
&= \lim_{x \to 0^{+}}\frac{10^{x} - 1}{x}\cdot\lim_{n \to \infty}\frac{\sqrt{n + 1} - \sqrt{n}}{1/\sqrt{n}}\cdot\lim_{y \to 0^{+}}\frac{y}{2^{y} - 1}\\
&= \frac{\log 10}{\log 2}\cdot\lim_{n \to \infty}\frac{\sqrt{n}}{\sqrt{n + 1} + \sqrt{n}}\\
&= \frac{\log 10}{\log 2}\cdot\lim_{n \to \infty}\dfrac{1}{\sqrt{1 + \dfrac{1}{n}} + 1}\\
&= \frac{\log 10}{2\log 2}\end{aligned}$$
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