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.

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 \(-\sin 3\).)

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

Different Interpretations:

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

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

\(3. \displaystyle \lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x} = \lim_{x \to 0} \frac{2 \sin x \cos4x}{\sin x} = 2\)

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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.

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-3

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Can you explain your thinking step by step?

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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 \(-\sin 3\).)

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