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yairl
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when i have lim x->0 (x-sinx)/(sinx)^3 why can't I replace it with 1/(sinx)^2-1/(sinx)^2?
thanks
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Replacing (x-sinx)/(sinx)^3 with 1/(sinx)^2-1/(sinx)^2 is not valid because they are not equivalent. The former expression simplifies to 1/sinx, while the latter simplifies to 0. This is because in the numerator of the first expression, the x and sinx terms cannot be combined, while in the second expression, the x and sinx terms cancel each other out.
No, you cannot use this rule in this scenario. This rule only applies when both the numerator and denominator functions have finite limits as x approaches the given value. In this case, both the numerator and denominator have a limit of 0, which is undefined.
Yes, there is a way to simplify the expression. By factoring out a sinx term from the numerator, we can rewrite the expression as (1-cosx)/sinx^2. This can then be simplified further to 1/sinx, which is equivalent to the original expression.
Yes, L'Hôpital's rule can be used to solve this limit. By taking the derivative of both the numerator and denominator, we can rewrite the expression as (1-cosx)/2sinx*cosx. This can then be simplified to 1/2cosx, which has a limit of 1/2 as x approaches 0.
The limit of (x-sinx)/(sinx)^3 as x approaches 0 is significant because it represents the slope of the tangent line to the curve y=sinx at the point (0,0). This is known as the derivative of sinx at x=0, and has a value of 1 at this point. This limit also plays a crucial role in the development of calculus and the study of functions and their behavior.