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

Question
Limits and continuity
Find the limits
$$\lim_{x\rightarrow0}\frac{\sin x}{2x^2-x}$$

2021-02-16
To evaluate: $$\lim_{x\rightarrow0}\frac{\sin x}{2x^2-x}$$
Solution:
$$\lim_{x\rightarrow0}\frac{\sin x}{2x^2-x}$$
On simplifying further, we get:
$$\lim_{x\rightarrow0}\frac{\sin x}{2x^2-x}=\lim_{x\rightarrow0}\frac{\sin x}{x(2x-1)}$$
$$=\lim_{x\rightarrow0}\frac{\sin x}{x}\times\frac{1}{(2x-1)}$$
$$=\lim_{x\rightarrow0}1\times\frac{1}{(2x-1)}$$
$$=\frac{1}{2(0)-1}$$
$$=\frac{1}{-1}$$
$$=-1$$
Result:
$$\lim_{x\rightarrow0}\frac{\sin x}{2x^2-x}=-1$$

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