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# Calculus

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

Note by Rishit Joshi
3 years, 10 months ago

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

Different Interpretations:

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

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

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

- 3 years, 10 months ago

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.

- 3 years, 10 months ago

-3

- 3 years, 10 months ago

Can you explain your thinking step by step?

Staff - 3 years, 10 months ago

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 $$-\sin 3$$.)

- 3 years, 10 months ago