Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ 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.
 A: 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 $$\lim_{x \to 0} \frac{\ln(x + \sqrt{1 + x^2}) - x}{\tan^3 x} = \frac{\left(-\frac{1}{6}\right)}{(1)} = -\frac{1}{6}.$$

A: 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 
$$\lim_{x \to 0}\dfrac{\ln[x + (1 + x^2)^{1/2}] - x}{\tan^3 x} = -\dfrac{1}{6}$$
A: 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*}
A: 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).
A: L'Hospital's rule:
$$\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{\tan^3(x)}=\lim_{x\to0} \frac{\ln(x+\sqrt{1+x^2})-x}{x^3}\cdot\frac{\tan^3 x}{x^3}=$$
$$=\lim_{x\to0} \frac{(\ln(x+\sqrt{1+x^2})-x)'}{(x^3)'}=\lim_{x\to0}\frac{\frac{1}{\sqrt{1+x^2}}-1}{3x^2}=\lim_{x\to0}\frac{-x^2}{3x^2\sqrt{1+x^2}(1+\sqrt{1+x^2})}=-\frac{1}{6}.$$
A: 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
A: 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.
