Limit of x-sinx/x^3 as x approaches 0

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In this video we find the limit of x-sinx/x^3 as x approaches 0.
Each step is explained carefully. We use L'Hopital's rule when required. Limit Laws are taken for granted in this problem.

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... At about time 1: 17 given: lim(x-->0)(sin(x)/6x) = (1/6)lim(x-->0)(sin(x)/x) = (1/6)(1) = 1/6 .Two other ways are: 1) by applying Taylor's theorem on numerator function and 2) Substitution x = 3T, and simplifying by applying the triple angle identity of sine (sin(3t) = 3sin(t) - 4sin^3(t)); this is a great strategy ... Thank you Alex and take care, Jan-W

jan-willemreens

Another approach here.
1. Let’s assume that lim (sin x - x)/x^3 converges to L as x->0.
2. Substitute x=3y, then (sin x - x)/x^3 = (sin (3y) - 3y)/(27*y^3).
We are going to use this lemma, sin(3y) = 3*sin y - 4*(sin y)^3.
Thus, (sin (3y) - 3y)/(27*y^3) = ((3*sin y - 3y) - 4*(sin y)^3)/(27*y^3) = 3*(sin y - y)/(27*y^3) - (4*(sin y)^3)/(27*y^3) = (sin y - y)/(9*y^3) - (4/27)*((sin y)^3)/(y^3)
3. We know that x->0 is same as y->0. Since lim (sin x - x)/x^3 as x->0 converges to L, lim (sin y - y)/(9*y^3) as y->0 converges to L/9. lim (4/27)*((sin y)^3)/(y^3) as y->0 convergence to 4/27.
4. Thus L=L/9 - 4/27. L = -1/6.
This is one of the ways to find lim (sin x - x)/x^3 as x->0 converges to -1/6. But this is not the proof yet. Because we didn’t prove lim (sin x - x)/x^3 converges as x->0.

DanielKang-tv

Can we find the limit without lhopitals rule

tumelofaithmphahlele

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