I was asked to solve the following limit using the nth Taylor polynomial with remainder.
$$
\displaystyle{\lim_{x \to 0}}\frac{\log{(1+x^2)}}{2x}.
$$
I couldn't find the remainder term because I couldn't generalize a formula for the nth derivative so I use the infinite series expansion
$$
-\sum_{k=1}^{\infty}\frac{(-1)^k(x^{2})^k}{k}.
$$
What would be the way to do it with the remainder?
Best Answer
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:
$$\lim_{x\to0}\frac {\log(1+x^2)}{2x} = \lim_{x\to0}\frac{o(x^2)}{2x}=0$$