<s>[INST] 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.
 [/INST] 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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