Using Maclaurin approximation to find the limit of $(\ln(1+x^2)-\ln(1-x^2))/(e^{x^2}-e^{-x^2})$ as $x\to 0$

Question

I have this assignment:

$$\lim_{x\to0}\frac{\ln(1+x^2)-\ln(1-x^2)}{e^{x^2}-e^{-x^2}}$$

And by using the Maclaurin approximations I get this:

$$\frac{x^2-\frac{x^4}{2}+x^2+\frac{x^4}{2}}{1+x^2+\frac{x^4}{2}-1+x^2-\frac{x^4}{2}}$$

which I can simplify to this:

$$\frac{2x^2}{2x^2}$$ and everything is right this far but then, in the solution they have just ignored the $x^2$ and written $\frac{2}{2} = 1$ How can it be so? I know that if the nominator and denominator is the same, then the answer would be 1, but in this case it would result in division by zero...

Answer 1

Let $l$ be a real number and let $f: x \mapsto f(x)$ be a real-valued function on the real field. Recall that "$f(x) \to l$ as $x \to 0$" means "$f(x)$ gets closer to $l$ as $x$ gets closer to $0$" and that $f$ need not be defined at $0$, and you know the rest.