= Solution
Combine the <logarithms> first and factor out the limiting constant:
$$
\ln(1+x+2x^2)-\ln x^2=\ln\left(2+\frac1x+\frac1{x^2}\right)
=\ln2+\ln\left(1+\frac1{2x}+\frac1{2x^2}\right).
$$
Apply the <Taylor expansion> $\ln(1+u)=u-u^2/2+O(u^3)$ with $u=1/(2x)+1/(2x^2)$. The first three nonzero terms, including the constant, are
$$
\boxed{\ln(1+x+2x^2)-\ln x^2=\ln2+\frac1{2x}+\frac3{8x^2}+O(x^{-3}),\qquad x\to\infty.}
$$
This is an <asymptotic expansion> in reciprocal powers of $x$.
Back to article page
