Solved: "Compute the limit: limx≥01−cos3(x)xsin⁡(2x)" - Plainmath

William Montgomery

Answered question

2022-01-26

Compute the limit:

lim_{x \geq 0} \frac{1 - \cos^{3} (x)}{x \sin (2 x)}

Answer & Explanation

Micah May

Beginner2022-01-27Added 11 answers

lim_{x \geq 0} \frac{1 - \cos^{3} (x)}{x \sin (2 x)}

Apply LHopitals Rule

lim_{x \to 0} \frac{1 - \cos^{3} x}{x \sin (2 x)} = lim_{x \to 0} \frac{3 \cos^{2} x \sin x}{\sin (2 x) + 2 x \cos (2 x)}

Again apply LHopitals Rule

lim_{x \to 0} \frac{3 (- \sin (2 x) \sin x + \cos^{3} x)}{4 \cos (2 x) - 4 x \sin (2 x)} = \frac{3}{4}

Darian Sosa

Beginner2022-01-28Added 12 answers

Hint: Write

lim_{x \to 0} \frac{1 - \cos^{3} x}{x \sin (2 x)} = lim_{x \to 0} \frac{1 - {(1 - \frac{x^{2}}{2})}^{3}}{2 x^{2}}

= lim_{x \to 0} \frac{1 - (1 - \frac{3 x^{2}}{2})}{2 x^{2}}

= lim_{x \to 0} \frac{{\frac{32}{x}}^{2}}{2 x^{2}}

= \frac{3}{4}

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