Answer
$\lim\limits_{x \to \ 0}\frac{x}{2x^2-x}=\lim\limits_{x \to \ 0}\frac{1}{2x-1}=\frac{1}{2(0)-1}=\frac{1}{-1}=-1$
Work Step by Step
If we want to calculate the value of $\lim\limits_{x \to \ 0}\frac{x}{2x^2-x}$,
Here, we cannot use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, because it would mean that:
$\lim\limits_{x \to \ 0}\frac{x}{2x^2-x}=\frac{0}{2(0)^2-0}=\frac{0}{0}$ which cannot be a limit.
Therefore, we have to simplify the function:
$\frac{x}{2x^2-x}=\frac{x}{x(2x-1)}=\frac{1}{2x-1}$
If we substitute this function into the limit, we get:
$\lim\limits_{x \to \ 0}\frac{x}{2x^2-x}=\lim\limits_{x \to \ 0}\frac{1}{2x-1}=\frac{1}{2(0)-1}=\frac{1}{-1}=-1$
