real analysis - Find $lim_{x to 0} frac{ln (x^2+1)} {x^2}$ without L'hopital's rule

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.

Answer

$$\begin{align} \lim_{x \to 0} \frac{\ln (x^2+1)} {x^2}&=\lim_{x \to 0} \ln (x^2+1)^{\frac{1}{x^2}}\\ &=\ln\left(\lim_{x \to 0} (x^2+1)^{\frac{1}{x^2}}\right)\\ &=\ln e=1 \end{align}$$