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Let $\ln(x+\sqrt{1+x^2})=:y$，then $x=\dfrac{1}{2}(e^y-e^{-y}).$ Therefore \begin{align*} \lim_{x \to 0}\left[\frac{\ln(x+\sqrt{1+x^2})}{x}\right]^{\frac{1}{x^2}}&=\lim_{y \to 0}\left(\frac{2ye^y}{e^{2y}-1}\right)^{\frac{4}{e^{2y}+e^{-2y}-2}}\\ &=\lim_{y \to 0}\left(1+\frac{2ye^y-e^{2y}+1}{e^{2y}-1}\right)^{\frac{e^{2y}-1}{2ye^y-e^{2y}+1}\cdot\frac{2ye^y-e^{2y}+1}{e^{2y}-1}\cdot\frac{4}{e^{2y}+e^{-2y}-2}}\\ &=\exp \lim_{y \to 0}\left(\frac{2ye^y-e^{2y}+1}{e^{2y}-1}\cdot\frac{4}{e^{2y}+e^{-2y}-2}\right)\\ &=\exp \lim_{y \to 0}\left(\frac{2ye^y-e^{2y}+1}{2y}\cdot\frac{4e^{2y}}{e^{4y}-2e^{2y}+1}\right)\\ &=\exp \left(2\lim_{y \to 0}\frac{2ye^y-e^{2y}+1}{ye^{4y}-2ye^{2y}+y}\right)\\ &=\exp \left[2\lim_{y \to 0}\frac{2e^y(y-e^y+1)}{(e^{2y}-1)(e^{2y}(4y+1)-1)}\right]\\ &=\exp \left[2\lim_{y \to 0}\frac{y-e^y+1}{e^{2y}(4y^2+y)-y}\right]\\ &=\exp \left[2\lim_{y \to 0}\frac{1-e^y}{e^{2y}(8y^2+10y+1)-1}\right]\\ &=\exp \left[-2\lim_{y \to 0}\frac{y}{e^{2y}(8y^2+10y+1)-1}\right]\\ &=\exp \left[-2\lim_{y \to 0}\frac{1}{e^{2y}(16y^2+36y+12)}\right]\\ &=\exp\left(-\frac{1}{6}\right). \end{align*}

Please correct me if I'm wrong! Are there simpler solutions?

hmakholm left over Monica
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mengdie1982
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    it is fine, however, you used three times L'Hospital and two times $\lim_\limits{x\to 0} \frac{e^x-1}{x}=1$, which need to be indicated for clarity. If that is the case, you can also change $x+\sqrt{1+x^2}=t$ or you can do it directly: $1+\frac{\ln (x+\sqrt{1+x^2})-x}{x}$ – farruhota May 24 '19 at 18:04

3 Answers3

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No need to substitute, Taylor expansion is enough. Denote the original limit by $L$. $$L=\exp\lim_{x\to0}\frac{\ln\frac{\ln(x+\sqrt{1+x^2})}{x}}{x^2}\\ =\exp\lim_{x\to0}\frac{\ln(1-\frac16x^2+O(x^3))}{x^2}\\ =\exp\lim_{x\to0}\frac{-\frac16x^2+O(x^3)}{x^2}\\ =\exp\left(-\frac16\right)$$ For the hint of expanding $\ln(x+\sqrt{1+x^2})$, try differentiate it and then expand.

Kemono Chen
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Note that $$\frac{d} {dx} \log\left(x+\sqrt{1+x^2}\right)=\dfrac{1+\dfrac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}$$ and hence we have $$\log\left(x+\sqrt{1+x^2}\right)=\int_{0}^{x}\frac{dt}{\sqrt{1+t^2}}$$ and therefore by Fundamental Theorem of Calculus $$f(x)=\dfrac{\log\left(x+\sqrt{1+x^2}\right)}{x}\to 1$$ If the desired limit is $L$ then by taking logs we can see that $$\log L=\lim_{x\to 0}\frac{\log f(x)} {x^2}$$ Since $f(x) \to 1$ we have $(\log f(x)) /(f(x) - 1)\to 1$ and we then have $$\log L=\lim_{x\to 0}\frac{f(x)-1}{x^2}=\lim_{x\to 0}\frac{\int_{0}^{x}\left\{(1+t^2)^{-1/2}-1\right\}\,dt}{x^3}$$ One easy option here is to apply L'Hospital's Rule to get the ratio $$\frac{(1+x^2)^{-1/2}-1}{3x^2}$$ which tends to $-1/6$ so that $L=e^{-1/6}$. Another alternative is to bound the integral suitably and apply Squeeze theorem.

Paramanand Singh
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Using $\ln(x + \sqrt{x^2 + 1}) = \sinh^{-1}(x) = x - \frac{x^3}{6} + \frac{3 x^5}{40} + \cdots$ then $$\ln\left(\frac{\sinh^{-1}(x)}{x}\right) = - \frac{x^2}{6} + \frac{11 x^4}{180} - \frac{191 x^6}{5670} + \cdots$$ which leads to \begin{align} \lim_{x \to 0} \left(\frac{\sinh^{-1}(x)}{x}\right)^{1/x^2} &= \lim_{x \to 0} e^{\frac{1}{x^2} \, (\ln(\sinh^{-1}(x)) - \ln(x))} \\ &= \lim_{x \to 0} e^{- \frac{1}{6} + \frac{11 x^2}{180} - \frac{191 x^4}{5670} + \cdots} \\ &= e^{-1/6} \end{align}

Leucippus
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