Find out the answer to "Find the limits limx→0sin⁡x2x2−x"

FizeauV

Answered question

2021-02-15

Find the limits

lim_{x \to 0} \frac{\sin x}{2 x^{2} - x}

Alannej

Skilled2021-02-16Added 104 answers

To evaluate:

lim_{x \to 0} \frac{\sin x}{2 x^{2} - x}

Solution:

lim_{x \to 0} \frac{\sin x}{2 x^{2} - x}

On simplifying further, we get:

lim_{x \to 0} \frac{\sin x}{2 x^{2} - x} = lim_{x \to 0} \frac{\sin x}{x (2 x - 1)}

= lim_{x \to 0} \frac{\sin x}{x} \times \frac{1}{(2 x - 1)}

= lim_{x \to 0} 1 \times \frac{1}{(2 x - 1)}

= \frac{1}{2 (0) - 1}

= \frac{1}{- 1}

= - 1

Result:

lim_{x \to 0} \frac{\sin x}{2 x^{2} - x} = - 1

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