lim 2x^3/(tan^3(2x)) x->0

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lim 2x^3/(tan^3(2x)) x->0

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Can anybody please help?
if u try draw the graph. u will get the limit is 1/4 when x->0
The thing is I need to prove this using identitities that (cosx-1)/x = 0 or sinx/x = 1

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Put the tanx in terms of cosx and sinx. Then it becomes cos(2x)^3* (2x^3/sin(2x)^3). You want the number inside the sin to be the same as the number on the numerator. Expanding the cube into the 2x you get sin8x^3. So you want the top to be 8x^3, so you have to multiple the top by 4. But since you multiple the top by 4, you also have to divide by 4. Pull the 4 you just divided out. Now the sin part is in the sinx/x=1 form so that part just equals one and the limit you are left with is (1/4)*cos(2x) which plugging in x=0 yields 1/4

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