Answer
$2$
Work Step by Step
Consider Taylor series for $\cos x $ as follows: $\cos x=1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....$
$\lim\limits_{x \to 0} \dfrac{\ln (1+x^2)}{1-\cos x}=\lim\limits_{x \to 0} \dfrac{(x^2-x^4/2+x^6/3)}{1-(1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....)}\\ \dfrac{\lim\limits_{x \to 0} (x^2-x^4/2+x^6/3)}{\lim\limits_{x \to 0}1-(1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....)}\\= \dfrac{\lim\limits_{x \to 0}(x^2-x^4/2+x^6/3)}{\lim\limits_{x \to 0}[1-(1-\dfrac{ x^2}{2!}+\dfrac{x^4}{4!}-....)]}\\=2 !$
or, $\lim\limits_{x \to 0} \dfrac{\ln (1+x^2)}{1-\cos x}=2$
