# Find the limits lim_{xrightarrow0}frac{sin x}{2x^2-x}

Find the limits
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2{x}^{2}-x}$
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Alannej
To evaluate: $\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2{x}^{2}-x}$
Solution:
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2{x}^{2}-x}$
On simplifying further, we get:
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2{x}^{2}-x}=\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x\left(2x-1\right)}$
$=\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}×\frac{1}{\left(2x-1\right)}$
$=\underset{x\to 0}{lim}1×\frac{1}{\left(2x-1\right)}$
$=\frac{1}{2\left(0\right)-1}$
$=\frac{1}{-1}$
$=-1$
Result:
$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2{x}^{2}-x}=-1$
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