anonymous
  • anonymous
lim 2x^3/(tan^3(2x)) x->0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Can anybody please help?
anonymous
  • anonymous
if u try draw the graph. u will get the limit is 1/4 when x->0
anonymous
  • anonymous
The thing is I need to prove this using identitities that (cosx-1)/x = 0 or sinx/x = 1

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anonymous
  • anonymous
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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