How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$

Perhaps not an elegant proof but I considered this.
Use a Taylor series for $\sin$.
$\sin(x)=x+ a_1x^3 + a_2x^5 + ...$
Let $A$ be the value of the limit.
Then we get
 $A=\dfrac{x^2+\frac{1}{x}-\frac{1}{x}+a_1(x^2+\frac{1}{x})^3-a_1(\frac{1}{x})^3+\ldots}{x} = \dfrac{x^2+a_1(x^2+\frac{1}{x})^3-a_1(\frac{1}{x})^3)+\ldots}{x}$.
This equals $A =\dfrac{x^2}{x}+\dfrac{a_1((x^3+1)^3-1)}{x^4}+\dfrac{a_2((x^3+1)^5-1)}{x^6}+\ldots$
Now use big $O$ notation to rewrite as follows : $A=\dfrac{O(x^2)}{x}+\dfrac{a_1O(x^9)}{x^4}+\dfrac{a_2O(x^{15})}{x^6}+\ldots$ hence $A=0 + a_1 0+a_20 + \ldots =0$. To justify that infinite sum notice that each limit $O(x^a)/x^b$ goes faster to $0$ than $x$ does.
