# How do you find the limit of sinx/(2x^2-x) as x approaches 0?

Jan 25, 2016

$- 1$

#### Explanation:

One thing you should know going into this is that

${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

This is an important identity in limit problems.

Keeping this in mind, we can factor an $x$ from the denominator of the fraction, giving

=lim_(xrarr0)(sinx)/(x(2x-1)

We can rearrange this to get $\sin \frac{x}{x}$, which we already know the limit of.

$= {\lim}_{x \rightarrow 0} \left(\sin \frac{x}{x}\right) \frac{1}{2 x - 1}$

Limits can be multiplied, as follows:

$= {\lim}_{x \rightarrow 0} \sin \frac{x}{x} \cdot {\lim}_{x \rightarrow 0} \frac{1}{2 x - 1}$

Since ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$, the expression simply equals

$= {\lim}_{x \rightarrow 0} \frac{1}{2 x - 1}$

And here, we can evaluate the limit straightaway by plugging in $0$ for $x$.

$= \frac{1}{2 \left(0\right) - 1} = - 1$

We can check the graph at $x = 0$:

graph{sinx/(2x^2-x) [-5.206, 5.89, -2.9, 2.65]}