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Evaluate the given limit:

$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$

I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show me what is the trick here?

Thanks.

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  • $\begingroup$ You will need l'Hospital 3 times. $\endgroup$ – orion Jan 21 '15 at 11:16
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    $\begingroup$ $\ln(x + \sqrt{1+x^2}) = \sinh^{-1} x$ $\endgroup$ – mvw Jan 21 '15 at 11:19
  • $\begingroup$ what do you exactly get confused with? show what you did so that you get help exactly on the issue you do not understand. $\endgroup$ – Math-fun Jan 21 '15 at 13:51
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Recall that $$\text{arsinh } x := \ln\left(x + \sqrt{1 + x^2}\right),$$ and that it has a nice Taylor series expansion: $$\text{arsinh } x \sim x - \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^5}{5} - \cdots$$ (it's not too hard to write the coefficients in a closed form, but we only need the first terms here).

Accounting for the subtracted $x$ in the numerator of the original ratio, the Taylor series of the numerator is $$- \frac{1}{6} x^3 + O(x^5).$$

Now, the Taylor series of $\tan x$ is$$\tan x \sim x + \frac{1}{3} x^3 + \frac{2}{15} x^5 + \cdots.$$ So, multiplying gives that the Taylor series of the denominator $\tan^3 x$ is $$x^3 + O(x^5).$$ The leading terms are both third-order, so the limit is the ratio of the coefficients of those terms, that is $$\lim_{x \to 0} \frac{\ln(x + \sqrt{1 + x^2}) - x}{\tan^3 x} = \frac{\left(-\frac{1}{6}\right)}{(1)} = -\frac{1}{6}.$$

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i will use the maclaurin expansion for $\sqrt{1+x}, \ln(1+x), \tan(x)$

$\begin{align} \ln[x + (1 + x^2)^{1/2}] &= \ln[x + 1 + \frac{1}{2}x^2 -\frac{1}{8}x^4 + \cdots]\\ &=\ln(1 + x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots) \\ &=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{1}{2} \{x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots\}^2 \\ &+ \frac{1}{3}\{x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots\}^3 \cdots\\ &=(x + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \cdots)-\frac{x^2}{2} (1 + x + \cdots) + \frac{1}{3}(x^3 + \cdots) +\cdots\\ &=x-\frac{1}{2}x^3 + \frac{1}{3}x^3 + \cdots\\ &= x-\frac{1}{6}x^3 + \cdots \end{align}$

the expansion for $$\tan x = x + \cdots$$ putting the two together

$$\lim_{x \to 0}\dfrac{\ln[x + (1 + x^2)^{1/2}] - x}{\tan^3 x} = -\dfrac{1}{6}$$

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This solution is based on no l'Hospital rule nor Taylor expansion. The following standard limits only are used:

\begin{eqnarray*} \lim_{x\rightarrow 0}e^{x} &=&1. \\ \lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1. \\ \lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}x^{2}-e^{x}}{x^{3}} &=&-\frac{1}{6}% . \\ \lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1-\frac{1}{2}x^{2}}{x^{3}} &=&0. \end{eqnarray*}

We transform the original expression $f(x)=\dfrac{\ln (x+\sqrt{1+x^{2}})-x}{\tan ^{3}x}$ as follows \begin{eqnarray*} f(x)&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (x+\sqrt{1+x^{2}})-\ln e^{x}}{% x^{3}} \\ &=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (\dfrac{x+\sqrt{1+x^{2}}}{e^{x}})}{% x^{3}} \\ &=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (1+\color{red}{u(x)})}{\color{red}{u(x)}}\cdot \dfrac{\color{red}{\dfrac{% x+\sqrt{1+x^{2}}-e^{x}}{e^{x}}}}{x^{3}},\ \ \ with\ \color{red}{u(x)=\frac{x+\sqrt{1+x^{2}}-e^{x}}{e^{x}}} \\ &=&\left( \frac{x}{\tan x}\right) ^{3}\cdot \dfrac{\ln (1+u(x))}{u(x)}\cdot \dfrac{1}{e^{x}}\cdot \left( \left( \frac{\sqrt{1+x^{2}}\color{green}{-1-\frac{1}{2}x^{2}}}{% x^{3}}\right) +\left( \frac{\color{blue}{1}+x\color{blue}{+\frac{1}{2}x^{2}}-e^{x}}{x^{3}}\right) \right) \end{eqnarray*} Since $\lim\limits_{x\rightarrow 0}u(x)=0$, and the function $t\rightarrow t^3$ is continuous, then \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\ln (x+\sqrt{1+x^{2}})-x}{\tan ^{3}x} &=&\left( \lim_{x\rightarrow 0}\frac{x}{\tan x}\right) ^{3}\cdot \lim_{u\rightarrow 0}% \frac{\ln (1+ u)}{u}\cdot \lim_{x\rightarrow 0}\frac{1}{e^{x}} \\ &&\cdot \left( \left( \lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1-\frac{1}{2}% x^{2}}{x^{3}}\right) +\left( \lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}% x^{2}-e^{x}}{x^{3}}\right) \right) \\ &=&\left( 1\right) ^{3}\cdot 1\cdot \frac{1}{1}\cdot \left( \left( 0\right) +\left( -\frac{1}{6}\right) \right) \\ &=&\color{red}{-\frac{1}{6}}. \end{eqnarray*}

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  • $\begingroup$ Hi, the third and fourth 'standard limits' used here require division by 0 , taking the limit of x^3 as x approaches 0. How are these two limits calculated? Thanks. $\endgroup$ – NetUser5y62 Sep 10 '17 at 6:15
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L'Hospital's rule: $$\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{\tan^3(x)}=\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{x^3}\cdot\frac{\tan^3 x}{x^3}=$$ $$=\lim_{x\to0} \frac{(\ln(x+\sqrt{1+x^2})-x)'}{(x^3)'}=\lim_{x\to0}\frac{\frac{1}{\sqrt{1+x^2}}-1}{3x^2}=\lim_{x\to0}\frac{-x^2}{3x^2\sqrt{1+x^2}(1+\sqrt{1+x^2})}=-\frac{1}{6}.$$

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Without Taylor, applying L'Hospital once:

The derivative of the numerator is $$\frac1{\sqrt{x^2+1}}-1=\frac{1-\sqrt{x^2+1}}{1+\sqrt{x^2+1}}\cdot\frac{1+\sqrt{x^2+1}}{\sqrt{x^2+1}}=-\frac{x^2}{(1+\sqrt{x^2+1})\sqrt{x^2+1}}.$$ Taking away the factors that tend to one, it can be simplified as $$-\frac{x^2}2.$$

The derivative of the denominator is $$3(\tan x)'\tan^2 x=3(\tan^2x+1)\tan^2x,$$ It can be simplified as $$3\tan^2x.$$

Now the limit is that of $$-\frac{x^2}{6\tan^2x}=-\frac16\left(\frac {\cos x}{\sin x}x\right)^2=-\frac16\cos^2x\left(\frac x{\sin x}\right)^2,$$ hence $$-\frac16.$$ Note: if you distrust the simplifications, you can keep all factors and split the limit as the product of two, one indeterminate (zero factors) and the other determinate (unit factors).

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Write $\tan x=\frac{\sin x}{\cos x}$

Hence it will be in $\frac{0}{0}$ form and then apply L'Hospital's rule 3 times

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The best approach here seems to be put $t = \log(x + \sqrt{1 + x^{2}})$ so that $$x + \sqrt{1 + x^{2}} = e^{t}, \sqrt{1 + x^{2}} - x = e^{-t}$$ and finally $$x = \frac{e^{t} - e^{-t}}{2} = \sinh t$$ Further note that $$\lim_{t \to 0}\frac{\sinh t}{t} = \lim_{t \to 0}\frac{e^{t} - e^{-t}}{2t} = \frac{1}{2}\lim_{t \to 0}\left(\frac{e^{t} - 1}{t} + \frac{e^{-t} - 1}{-t}\right) = 1$$We can now proceed as follows \begin{align} L &= \lim_{x \to 0}\frac{\log(x + \sqrt{1 + x^{2}}) - x}{\tan^{3}x}\notag\\ &= \lim_{x \to 0}\frac{\log(x + \sqrt{1 + x^{2}}) - x}{x^{3}}\cdot\frac{x^{3}}{\tan^{3}x}\notag\\ &= \lim_{x \to 0}\frac{\log(x + \sqrt{1 + x^{2}}) - x}{x^{3}}\notag\\ &= \lim_{t \to 0}\frac{t - \sinh t}{\sinh^{3}t}\text{ (putting }x = \sinh t)\notag\\ &= \lim_{t \to 0}\frac{t - \sinh t}{t^{3}}\cdot\frac{t^{3}}{\sinh^{3}t}\notag\\ &= \lim_{t \to 0}\frac{t - \sinh t}{t^{3}}\tag{1}\\ &= \lim_{t \to 0}\frac{1 - \cosh t}{3t^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{3}\lim_{t \to 0}\frac{-2\sinh^{2}(t/2)}{t^{2}}\notag\\ &= -\frac{2}{3}\lim_{t \to 0}\frac{\sinh^{2}(t/2)}{4(t/2)^{2}}\notag\\ &= -\frac{1}{6}\notag\\ &= \text{(continue from equation (1))}\notag\\ &= \lim_{t \to 0}\dfrac{t - \left(t + \dfrac{t^{3}}{6} + o(t^{3})\right)}{t^{3}}\text{ (using Taylor series for }\sinh t)\notag\\ &= -\frac{1}{6}\notag \end{align} The problem can't be solved without use of LHR or Taylor. Moreover if we go for LHR, only one application is sufficient.

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