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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    Can anybody please help?

  2. anonymous
    • 5 years ago
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    if u try draw the graph. u will get the limit is 1/4 when x->0

  3. anonymous
    • 5 years ago
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    The thing is I need to prove this using identitities that (cosx-1)/x = 0 or sinx/x = 1

  4. anonymous
    • 5 years ago
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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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