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Calculus

LIMIT X TENDS TO 0 SIN5X-SIN3X/SINX???

Note by Rishit Joshi
3 years, 5 months ago

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Hi, rishit, firstly, you should try to learn latex here

Different Interpretations:

\(1. \displaystyle \lim_{x \to 0} \sin 5x - \sin \bigg(\frac{3x}{\sin x}\bigg) = -\sin 3 \)

\(2.\displaystyle \lim _{x \to 0} \sin 5x - \frac{\sin 3x}{\sin x} = -3\)

\(3. \displaystyle \lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x} = \lim_{x \to 0} \frac{2 \sin x \cos4x}{\sin x} = 2\) Jatin Yadav · 3 years, 5 months ago

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Using L' Hopital's Rule, we take the derivative of the numerator and denominator of the function so we have (5 cos(5x) - 3 cos(3x)) / cos x, this is the time to take the limit and simply substitute the value... so this limit as the function tends to zero is 2. John Ashley Capellan · 3 years, 5 months ago

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-3 Vinayak Bodake · 3 years, 5 months ago

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@Vinayak Bodake Can you explain your thinking step by step? Calvin Lin Staff · 3 years, 5 months ago

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@Calvin Lin Obviously this person considers the expression as \(\sin 5x - \frac{\sin 3x}{\sin x}\), which is a perfectly valid interpretation. In fact, the one with \(\frac{\sin 5x - \sin 3x}{\sin x}\) is wrong since it would be (sin 5x - sin 3x)/sin x in linear form; note the parentheses.

(Moreover, I believe the correct interpretation according to how the problem is currently phrased is thus \(\sin 5x - \sin \frac{3x}{\sin x}\). In this case it seems that the limit is \(-\sin 3\).) Ivan Koswara · 3 years, 5 months ago

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