Some standard limits and property of logarithms,
$1)\lim_{x \to 0} \dfrac{\tan x}{x} = 1$
$2)\lim_{x \to 0} \dfrac{\ln (x + 1) }{x} = 1$
$3) \ln (ab) = \ln a + \ln b$
Given limit , $\lim_{x \to 0} \dfrac{\ln ((x^2 +1) \cos x)}{x^2}$ we can rewrite it as,
$\lim_{x \to 0} \left(\dfrac{\ln (x^2 +1)}{x^2} + \dfrac{\ln (\cos x)}{x^2}\right)$
now lets evaluate individually,
Let $y=x^2$ ,as $x \to 0$ $y \to 0$, for the limit below we can use substitution with $y=x^2$
$\lim_{x \to 0}\dfrac{\ln (x^2 +1)}{x^2}$ $=\lim_{y \to 0}\dfrac{\ln (y +1)}{y}$
now the limit is in form of the first standard limit which is mentioned above so,
$=\lim_{y \to 0}\dfrac{\ln (y +1)}{y}$$ =1$
second limit is, $\lim_{x \to 0}\dfrac{\ln (\cos x)}{x^2}$ $\left(\dfrac{0}{0} form \right)$, so apply L'Hopital rule,
=$\lim_{x \to 0}-\dfrac{\tan x }{2x}$ $= -\dfrac{1}{2}\lim_{x \to 0}\dfrac{\tan x }{x}$
(now its in form of second standard limit mentioned above so,)
$= -\dfrac{1}{2}\lim_{x \to 0}\dfrac{\tan x }{x}$$= -\dfrac{1}{2}(1)$ $= -\dfrac{1}{2}$
as both these limits exists we can use sum rule of limits as,
$\lim_{x \to 0} \left(\dfrac{\ln (x^2 +1)}{x^2} + \dfrac{\ln (\cos x)}{x^2}\right)$ $=\lim_{x \to 0}\dfrac{\ln (x^2 +1)}{x^2} + \lim_{x \to 0}\dfrac{\ln (\cos x)}{x^2}$$= 1 - \dfrac{1}{2}$$= \dfrac{1}{2}$.
