the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.

Various tricks can simplify the limit. Multiply by $\frac{\sin(x^2)}{x^2}$ which has a limit of one to reduce to
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^3}.$$Add $$\frac{\sin x^2 -x^2}{x^3}$$ to reduce to
$$\frac{(x^2+1) \ln(x^2+1)-x^2}{x^3}.$$Now break into pieces,
$$\frac{\ln(x^2+1)-1}{x}+\frac{\ln(x^2+1)}{x^3}.$$
