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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?

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2 Answers 2

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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:

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

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As an alternative by standard limit we have that

$$\frac{\log{(1+x^2)}}{2x}=\frac{\log{(1+x^2)}}{x^2}\frac{x^2}{2x} \to 1 \cdot 0 =0$$

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