Find this somewhat unpleasant limit

Question

I am solving a problem and after quite some computations and almost 1 day spent on it, I decided to ask here.

It all boils down to finding this limit $$\lim_{x \to 0+}C(x) = ?$$

where

$$C(x) = \frac{x-\sqrt{x^2+1}\cdot \ln \big(\sqrt{x^2+1} + x \big)}{2x^2\sqrt{x^2+1}\cdot \ln\big(\sqrt{x^2+1} + x \big)}$$

I applied L'Hôpital's rule a few times to get to here.

Now WA says this limit is $$-1/6$$ and this is correct. So I am trying to compute this and derive that $$\lim_{x \to 0+}C(x) = -1/6$$ by hand.

And after studying the sub-expressions, I can see this limit is of the kind $0/0$ but if I try to apply L'Hôpital's rule again to the expression $C(x)$, it doesn't get simpler, it gets more complicated.

So there must be some trick here. Maybe I need to divide the numerator and denominator by some expression. I tried that too a few times but I don't succeed at making it simpler.

Or... is this problem not solvable at all just by using L'Hôpital's rule?
But I don't see what other theory to apply here.

Any help or hint as to how to proceed?

Side note: Here is the original problem which led me to this expression $C(x)$. It asks us to find this limit.

$$\lim_{x \to 0+} \left(\frac{\ln(x+\sqrt{1+x^2})}{x}\right)^\frac{1}{x^2} = ?$$

Answer 1

A possible way to "see" the limit is the following. By Taylor's theorem we have $$ \log(x + \sqrt{1+x^2}) = x - \frac{x^3}{6} + R_5(x) $$ with a remainder $R_5$.

Therefore, $$ \left(\frac{\log(x + \sqrt{1+x^2})}{x} \right)^{1/x^2} = \left(1 - x^2\left(\frac{1}{6} - \frac{R_5(x)}{x^3} \right) \right)^{1/x^2}. $$

The remainder term $\frac{R_5(x)}{x^3}$ can be bounded by the usual techniques (see e.g. the wikipedia site "Taylor's theorem"). With a sandwhiching argument we can conclude $$ \lim_{x \to 0+}{\left(\frac{\log(x + \sqrt{1+x^2})}{x} \right)^{1/x^2}} = \lim_{x \to 0+}{\left(1 - \frac{x^2}{6}\right)}^{1/x^2} = e^{-1/6}. $$

Answer 2

$\def\lm{\lim_{x\to0^+}} \def\sq{\sqrt{1+x^2}} \def\lg{\ln(x+\sq)} \def\lh{\textrm{l'Hopital's}}$We evaluate the limit, $\lm C(x)$, using l'Hopital's, showing all necessary details. When verifying the work below, it is useful to notice that \begin{align*} \frac{d}{dx}\lg &= \frac{1}{\sq} \\ \frac{d}{dx}\sq &= \frac{x}{\sq}. \end{align*} We find \begin{align*} \lm C(x) &= \lm \frac{x-\sq\lg}{2x^2\sq\lg} \\ &= \lm \frac{1-\frac{x}{\sq}\lg-\sq\frac{1}{\sq}}{4x\sq\lg+2x^2\frac{x}{\sq}\lg+2x^2\sq\frac{1}{\sq}} & \lh \\ &= \lm \frac{-\frac{x\lg}{\sq}}{\frac{4x(1+x^2)\lg}{\sq}+\frac{2x^3\lg}{\sq}+2x^2} \\ &= \lm \frac{-\frac{x\lg}{\sq}}{2x^2+\frac{(4x+6x^3)\lg}{\sq}} \\ &= \lm \frac{-\frac{x\lg}{\sq}}{2x^2+\frac{(4x+6x^3)\lg}{\sq}} \frac{\frac{\sq}{x}}{\frac{\sq}{x}} \\ &= \lm \frac{-\lg}{2x\sq+(4+6x^2)\lg} \\ &= \lm \frac{-\frac{1}{\sq}}{2\sq+2x\frac{x}{\sq}+12x\lg+(4+6x^2)\frac{1}{\sq}} & \lh \\ &= -\frac{1}{6}. \end{align*}