I'm trying to prove that:
$\dfrac{\sin(x)^2\log(x^2)}{x} \to 0$ as $x \to 0^+$
I have got to the following:
$\left| \dfrac{\sin(x)^2\log(x^2)}{x} - 0 \right| = \dfrac{|\sin(x)^2||2\log(x)|}{|x|} \leq \dfrac{|x|^2|2\log(x)|}{|x|}=|x||2\log(x)| $
I don't know how to proceed from here, or if this is the right/best way to start. I hope someone can tell me what im missing.
I think you can easily, changing $x$ by $\lfloor x\rfloor$, reduce the asked (write, if not) case to a fraction $\frac{\log _{2}n}{n}, n\in \mathbb{N}$ and let me show some technique for its estimation.
Let's firstly start with $\frac{n}{b^n}$, where $b>1$, and show, that for any $\forall \varepsilon>0$ we can find such $N$, that for any $n>N$ holds $\frac{n}{b^n}<\varepsilon$ (de facto this is limit definition). This easily comes from inequality $$\frac{n}{b^n} = \frac{n}{(1+b-1)^n}=\frac{n}{1+n(b-1)+\frac{n(n+1)}{2}(b-1)^2+\cdots + b^n}<\frac{n}{\frac{n(n+1)}{2}(b-1)^2}=\\=\frac{2}{(b-1)^2(n+1)}<\varepsilon$$ Solving extreme right inequality we obtain $n>\frac{2}{(b-1)^2\varepsilon}-1$, so for $N=\left\lfloor \frac{2}{(b-1)^2\varepsilon}-1\right\rfloor +1$ we get desired sentence.
Come back to logarithm. Using what has just been proven and taking $\varepsilon =1$ we can get such $N$, that for any $n>N$ holds $\frac{1}{b^n}<\frac{n}{b^n}<1$. Let's consider any $\forall \varepsilon_1>0$. We can consider $b=2^{\varepsilon_1}$ and have $\dfrac{1}{2^{n\varepsilon_1}}<\dfrac{n}{2^{n\varepsilon_1}}<1$. It is same with $1<n<2^{n\varepsilon_1}$. Taking logarithm from all sides, gives $0<\log_2 n< n\varepsilon_1$ or $$0<\frac{\log_2 n}{n}<\varepsilon_1\quad(1)$$ for all $n>N$.
Obtained can be extended to an any $0<\alpha<1$ by following
$\dfrac{\log _{2}n}{n^{\alpha}} = \dfrac{1}{\alpha} \cdot \dfrac{\log _{2}n^{\alpha}}{n^{\alpha}} \leqslant \dfrac{1}{\alpha} \cdot \dfrac{\log _{2}(\lfloor n^{\alpha} \rfloor + 1)}{\lfloor n^{\alpha} \rfloor} = \dfrac{1}{\alpha} \cdot \dfrac{\log _{2}\lfloor n^{\alpha} \rfloor }{\lfloor n^{\alpha} \rfloor} + \dfrac{1}{\alpha} \cdot \dfrac{\log _{2}(1+\frac{1}{\lfloor n^{\alpha} \rfloor})}{\lfloor n^{\alpha} \rfloor}$ First summand is subsequense of $(1)$ and second can be estimated by constant multiply on $\frac{1}{\lfloor n^{\alpha} \rfloor}$. This can be taken as exact basis for $\lim\limits_{x\to 0}x^{\alpha}\log x =0$.