How to calculate $\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}$?

You can't substitute a function with an equivalent if sums are involved. Here it's better to use Taylor expansion: $$ \ln(1+x)=x-\frac{x^2}{2}+o(x^2) $$ so your limit becomes $$ \lim_{x\to0}\frac{(x-x^2/2)+(-x-x^2/2)+o(x^2)}{x^2} $$

As you see, $x$ and $-x$ cancel out, but there's something of the order of $x^2$ remaining.

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Comment

You could use equivalents by noticing that $\ln(1+x)+\ln(1-x)=\ln(1-x^2)$, which is equivalent to $-x^2$, but this works in the particular case and would not help in a case such as $$ \lim_{x\to0}\frac{x-\ln(1+x)}{x^2} $$

Use

• $\ln(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3} - \cdots$

• $\ln(1-x) = -x -\frac{x^2}{2}-\frac{x^3}{3}-\cdots$

Then you have $$\ln(1+x)+\ln(1-x) = -x^{2} -\frac{x^4}{2} + \cdots =x^{2}\cdot \left(-1 -\frac{x^2}{2} + \cdots\right)$$

$$\begin{align}\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2} &= \lim_{x\to0} \frac{-\ln(1-x^2)}{-x^2} = -1 \end{align}$$

Since $\lim_{x\to 0} \dfrac{\ln(x + 1)}{x} = 1$.