I am new at computing limits with infinitesimals and I am having trouble solving this one:
$$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}$$
I tried to substitute by means of equivalent infinitesimals and I came up with this:
$$\lim_{x\to0}\frac{x-x}{x^2}$$ But I do not know how to continue. The result must be $-1$. Any help would be appreciated!
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.
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 $\lim_{x\to 0} \dfrac{\ln(x + 1)}{x} = 1$.
We apply L'Hospital:
$f(x)=\ln(x+1)+\ln(1-x)$
$g(x)=x^2$
Note that $f(0)=g(0)=0$.
It is $f'(x)=\frac{1}{x+1}+\frac{1}{x-1}$ and $g'(x)=2x$. Note that $f'(0)=g'(0)=0$ so we apply L'Hospital a second time:
$f'(x)=-\frac{1}{(x+1)^2}-\frac{1}{(1-x)^2}$ and $g''(x)=2$
With $f''(0)=-2$ and $g''(0)=2$. Hence:
$\lim_{x\to 0} \frac{\ln(x+1)+\ln(1-x)}{x^2}=\frac{-2}{2}=-1$
By L'Hopital Rule we have $$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2} = \lim_{x\to0} \frac{\frac{1}{1+x}-\frac{1}{1-x}}{2x} =\lim_{x\to0} \frac{\frac{-2x}{1-x^2}}{2x} =\lim_{x\to0} \frac{-1}{1-x^2}= -1$$
$$\lim_{x\to0} \frac{\ln(1+x)+\ln(1-x)}{x^2}=\lim_{x\to0} \ln(1-x^2)^{1/x^2}=\lim_{t\to\infty} -\ln(1-\dfrac1t)^{-t}=-1$$
