Taylor polynomial with remainder for solving limit

Let $f(x)=\log(1+x^2)$.

Then $f'(x)=\dfrac {2x}{1+x^2}$.

This gives $\log (1+x^2) = f(0)+f'(0)x+o(x^2) = o(x^2)$.

(This $o(x^2)$ is the remainder term, which is $\dfrac {f''(\xi)}{2!}x^2$ in Lagrange form.)

Thus we have:

$$\lim_{x\to0}\frac {\log(1+x^2)}{2x} = \lim_{x\to0}\frac{o(x^2)}{2x}=0$$