How to Prove $\lim\limits_{k\to0}{(\frac{1}{kv}x+\frac{1}{k^2}\ln(1-\frac{kx}{v}))}=-\frac{x^2}{2v^2}$ How do you prove this equality: $\lim\limits_{k\to0}{(\frac{1}{kv}x+\frac{1}{k^2}\ln(1-\frac{kx}{v}))}=-\frac{x^2}{2v^2}$ ? I'm especially surprised by the appearance of $x^2$ on the R.H.S., but this equality seems legitimate.

By Taylor expansion of the logarithm, we obtain
$$\begin{align}
\lim_{k\to 0} \left(\frac{1}{kv}x + \frac{1}{k^2}\ln \left(1 - \frac{kx}{v}\right)\right) &= \lim_{k\to 0} \left(\frac{1}{kv}x + \frac{1}{k^2}\left(-\frac{kx}{v} - \frac{k^2x^2}{2v^2} + O(k^3)\right)\right)\\
&= \lim_{k\to 0}\left(-\frac{x^2}{2v^2} + O(k)\right)\\
&= -\frac{x^2}{2v^2}.
\end{align}$$
