= Solution
Near zero, the <Taylor expansions> are
$$
\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4),\qquad \sqrt{1+2x}=1+x-\frac{x^2}{2}+\frac{x^3}{2}+O(x^4).
$$
The constant, linear and quadratic terms cancel in the numerator. Consequently
$$
1+\ln(1+x)-\sqrt{1+2x}=-\frac{x^3}{6}+O(x^4),\qquad \boxed{\lim_{x\to0}\frac{1+\ln(1+x)-\sqrt{1+2x}}{x^3}=-\frac16.}
$$
The square root and logarithm are both real in a sufficiently small neighborhood of zero, so the two-sided real limit is well defined.
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