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.
 A: Just to suggest another way: $\ln(1+x^2)=x^2-\frac{x^4}2+O(x^6)$ and $\sin(x^2)=x^2+O(x^6)$, so the numerator behaves like $(x^2+1)(x^2-\frac{x^4}2)-x^2\sim\frac{x^4}2$. The denominator on the other hand behaves like $x^3$.

Proof that the limit is $0$.

*

*By simple study of functions, we check that $\lvert\ln(x^2+1)-x^2\rvert\le\frac{x^4}2$ and $\lvert\sin(x^2)-x^2\rvert\le\frac{x^6}6$.

*Then we can bound the numerator as
\begin{align*}
\Bigl|(x^2+1)\ln(x^2+1)-\sin(x^2)\Bigr|&\le x^2\ln(x^2+1)+\lvert\ln(x^2+1)-x^2\rvert+|x^2-\sin(x^2)|\\
&\le x^4+\frac{x^6}2+\frac{x^4}2+\frac{x^6}2\\
&=\frac32x^4+x^6.
\end{align*}

*Now
$$\left\lvert\frac{(x^2+1)\ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}\right\rvert
\le\frac{\frac32x^2+x^4}{|x|}\cdot\frac{x^2}{\lvert\sin(x^2)\rvert}
=\left(\frac32|x|+|x|^3\right)\left\lvert\frac{x^2}{\sin(x^2)}\right\rvert.$$
A: 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,
$$\frac{\ln(x^2+1)-1}{x}+\frac{\ln(x^2+1)}{x^3}.$$
A: 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 $x$.
