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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 $\frac{1}{2}+\frac{\ln{5}}{\ln{4}}$

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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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Hint $$ a^x-1=e^{x\ln a}-1= x\ln a+o(x) $$

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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

$$\displaystyle \lim_{n\to \infty} {\frac{10^{\sqrt{n+1}-\sqrt{n}}-1}{2^{\frac{1}{\sqrt{n}}}-1}}=\lim_{x\to 0^+} \frac{10^{\frac{x}{1+\sqrt{x^2+1}}}-1}{2^x-1} \underbrace{=}_{\mathrm{L'Hospital}} \lim_{x\to 0^+} \frac{10^{\frac{x}{1+\sqrt{x^2+1}}} \frac{1}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} \ln 10}{2^x\ln 2}\\=\frac{\frac12 \ln 10}{\ln 2}\underbrace{=}_{\ln 10=\ln (2\cdot 5)=\ln 2+\ln 5}\frac{\frac12 \ln 2+\frac 12 \ln 5}{\ln 2}=\frac12+\frac{\ln 5}{2\ln 2}=\frac12+\frac{\ln 5}{\ln 2^2}=\frac12+\frac{\ln 5}{\ln 4}.$$

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