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

FizeauV 2021-02-15 Answered
Find the limits
limx0sinx2x2x
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Expert Answer

Alannej
Answered 2021-02-16 Author has 104 answers
To evaluate: limx0sinx2x2x
Solution:
limx0sinx2x2x
On simplifying further, we get:
limx0sinx2x2x=limx0sinxx(2x1)
=limx0sinxx×1(2x1)
=limx01×1(2x1)
=12(0)1
=11
=1
Result:
limx0sinx2x2x=1
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