Find $\lim_{x\to 0}\frac{|\sin x|}{x}$ Can I say that $$\lim_{x\to 0}\frac{\sin x}{x}=1 \implies \lim_{x\to 0}\frac{|\sin x|}{x}=1$$?
Exercise: show that 
$$
\lim_{x\to 0}\frac{|\sin x|}{x}
$$ does not exists.
I know three ways to attack this. The sandwich theorem, L'Hospital's Rule or Taylor series.
I will show you the way with the Taylor series. Let's look at the series expansion of $\sin{x}$ at the point $0$:
$$\sin{x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...
$$
Now you can see that
$$
\frac{\sin{x}}{x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...
$$
Taking the limit now will be trivial if you belief in Taylor series. Maybe you can proof the series expansion of $\sin{x}$.