TITLE: Find $\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2} $ without L'hopital's rule
QUESTION [5 upvotes]: I have to find the limit without L'hopital's rule:
$$\lim_{x \to 0} \frac{\ln (x^2+1)} {x^2} $$
Is it possible? 
I thought about using squeeze theorem or something, but it didn't work out. 
Hints are more than welcome! 
P.S -  I didn't study Taylor series or Integrals yet.

REPLY [3 votes]: With series:
$ \ln(x^2+1)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-+...$ for $|x|<1$,
hence 
$\frac{\ln (x^2+1)} {x^2}=1-\frac{x^2}{2}+\frac{x^4}{3}-+... \to 1$ for $x \to 0$