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5 February, 08:04

Limit as x approaches 0 of. (sin3xsin5x) / x^2

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  1. 5 February, 09:53
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    The limit of the function (sin3x sin5x) / x^2 as x approaches zero is evaulated by substituting the function by zero. Since the answer is zero / zero which is indeterminate. Using L'hopitals rule, we derive separately the numerator and the denominator. we all know that sin 5x and sin 3x are equal to zero. Upon teh first derivative, the answer is still zero / zero. We derive further until the function has a denominator of 2 and a numerator still equal to zero. Since the answer is now zero / 2 or zero not zero/zero, the limit then is equal to zero.
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