Question

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

Limits and continuity
ANSWERED
asked 2021-02-15
Find the limits
\(\lim_{x\rightarrow0}\frac{\sin x}{2x^2-x}\)

Answers (1)

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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