I have to find the limit without L'hopital's rule:
lim
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}
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