= Solution
Both numerator and denominator vanish quadratically. Factor out $x^2$ before dividing the <power series>:
$$
\frac{x\sin x}{\ln(1+x^2)}
=\frac{1-x^2/6+x^4/120+O(x^6)}{1-x^2/2+x^4/3+O(x^6)}.
$$
The reciprocal of the denominator is $1+x^2/2-x^4/12+O(x^6)$. Hence
$$
\boxed{\frac{x\sin x}{\ln(1+x^2)}
=1+\frac{x^2}{3}-\frac{19x^4}{120}+O(x^6).}
$$
The expansion describes the continuous extension at the <removable singularity> $x=0$, where the original quotient is undefined but its limit is one. Odd powers vanish because the quotient is even.
Back to article page
