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.

Make life easier writing
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}=x\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^2\sin(x^2)}$$ Now, let $t=x^2$ to make
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^2\sin(x^2)}=\frac{(t+1) \ln(t+1)-\sin(t)}{t\sin(t)}$$ Using equivalent $\ln(t+1)\sim t$ and $\sin(t)\sim t$
$$\frac{(t+1) \ln(t+1)-\sin(t)}{t\sin(t)}\sim \frac{(t+1)t-t }{t^2 }=1$$
So, the expression is "similar" to $x$.
