Consider the problem, $\displaystyle \lim_{x\to\0}\frac{sinx}{2x^2-x}=-1$.

I tried u-substitution; let $\displaystyle u=2x^2-x$, then $\displaystyle \lim_{x\to\0}u=0$;

for, $\displaystyle (u=2x^2-x)$ $\displaystyle \vee$ $\displaystyle (\lim_{x\to\0}2x^2-x=\lim_{x\to\0}2(0)^2-0=0)$$\displaystyle \subset$ $\displaystyle \lim_{x\to\0}u=0$.

From this,

$\displaystyle \lim_{x\to\0}\frac{sinx}{2x^2-x}=\frac{\lim_{x\to\0}sinx}{\lim_{u\to\0}u}$ is effectively, $\displaystyle \lim_{x\to\0}\frac{sinx}{x}=1$

I figured I could treat the limits for the numerator and denominator independently as it doesn't seem to violate any limit laws—is this okay?

The question: where did the negative sign arise from?