"Evaluating trigonometric limit csc2(2x)−14x2" was answered by our experts

Kendall Daniels

Answered question

2022-04-02

Evaluating trigonometric limit

\csc^{2} (2 x) - \frac{1}{4 x^{2}}

arrebusnaavbz

Beginner2022-04-03Added 18 answers

Based on DeepSea's hint, I've managed to solve by myself.

lim_{x \to 0} [\csc^{2} (2 x) - \frac{1}{4 x^{2}}]

= lim_{x \to 0} [\frac{1}{\sin^{2} (2 x)} - \frac{1}{4 x^{2}}]

= lim_{x \to 0} [\frac{4 x^{2} - \sin^{2} (2 x)}{4 x^{2} (\sin^{2} (2 x))}]

= lim_{x \to 0} [\frac{4 x^{2} - \sin^{2} (2 x)}{4 x^{2} \times 4 x^{2} {(\frac{\sin (2 x)}{2 x})}^{2}}]

= lim_{x \to 0} [\frac{4 x^{2} - \sin^{2} (2 x)}{4 x^{2} \times 4 x^{2}}] \times lim_{x \to 0} [\frac{1}{{(\frac{\sin (2 x)}{2 x})}^{2}}]

= lim_{x \to 0} [\frac{4 x^{2} - \sin^{2} (2 x)}{16 x^{4}}] \times 1

= lim_{x \to 0} [\frac{8 x - 2 \sin (4 x)}{64 x^{3}}]

= lim_{x \to 0} [\frac{8 - 8 \cos (4 x)}{192 x^{2}}]

= lim_{x \to 0} [\frac{1 - \cos (4 x)}{24 x^{2}}]

= \frac{1}{24} \times lim_{x \to 0} [\frac{1 - \cos (4 x)}{x^{2}}]

= \frac{1}{24} \times \frac{1}{2} \times 4^{2}

= \frac{1}{3}

Lana Hamilton

Beginner2022-04-04Added 12 answers

hint: replace the denominator on the third line of your proof

\sin^{2} (2 x) by {(2 x)}^{2}

and apply L'hopitale rule three times to the expression:

\frac{4 x^{2} - \sin^{2} (2 x)}{16 x^{4}}

.I did it and it works. Try it. Note that "replace" here means you write:

\sin^{2} (2 x) = 4 x^{2} \cdot {(\frac{\sin (2 x)}{2 x})}^{2}

and the second factor approaches 1 when

x \to 0

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