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

Aug 4, 2016

-1

#### Explanation:

$\textcolor{b l u e}{{1}^{s t} \text{ Method}}$

First we check if we can use L'Hôpital's rule. For this to be viable we need the limit to be of indeterminate form.

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

So limit is indeterminate and we can use L'Hôpital's.

${\lim}_{x \rightarrow 0} \frac{\sin x}{2 {x}^{2} - x} = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} \left(\sin x\right)}{\frac{d}{\mathrm{dx}} \left(2 {x}^{2} - x\right)}$

$= {\lim}_{x \rightarrow 0} \frac{\cos x}{4 x - 1} = \frac{\cos \left(0\right)}{4 \left(0\right) - 1} = \frac{1}{- 1} = - 1$

$\textcolor{b l u e}{{2}^{n d} \text{ Method}}$

Rewrite limit as

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

It's a logical fallacy to use L'Hôpital's on the first term so we use alternative methods such as the Squeeze Theorem to show that

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

So overall limit becomes:

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