TrojanPoem
  • TrojanPoem
Find the limit of : ( No L'hopital's rule)
Mathematics
schrodinger
  • schrodinger
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TrojanPoem
  • TrojanPoem
\[\lim_{x \rightarrow \frac{ \pi }{ 6 }} \frac{ 2\sin x -1 }{ \sqrt{3} - 2\cos x }\]
TrojanPoem
  • TrojanPoem
Approaches ( pi/ 6 )
phi
  • phi
I would use L'hopital's rule

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TrojanPoem
  • TrojanPoem
Out of my curriculum.
phi
  • phi
one trick that often works is to multiply by the conjugate of the denominator the idea is (a+b)(a-b) = a^2 - b^2
TrojanPoem
  • TrojanPoem
But what made us think of doing so. To me It's clear when there is square roots.
TrojanPoem
  • TrojanPoem
I even never so example when they multiply by the conjugate with sin , cos , tan
TrojanPoem
  • TrojanPoem
i have never even seen*
phi
  • phi
I agree it is not an obvious tactic.
TrojanPoem
  • TrojanPoem
After talking to my teacher , he told me that It was included in exam in 1998 and showed me it , he even saw the answer and wasn't obvious to him.
TrojanPoem
  • TrojanPoem
was confused by it*
TrojanPoem
  • TrojanPoem
By the way , thanks phi.
TrojanPoem
  • TrojanPoem
If you ever though of trying to it and found alternative solution , Pm me !
phi
  • phi
multiplying top and bottom by (2 sin x + 1)( sqr(3) + 2 cos x) doesn't seem to help (as far as I can see)
TrojanPoem
  • TrojanPoem
Give me a second , I will see the solution written in my notebook.
TrojanPoem
  • TrojanPoem
(4sin^2x - 1) (sqrt(3) + 2cosx) / (3 - 4 cos ^2 x) (2 sinx +1 )
TrojanPoem
  • TrojanPoem
(4( 1- cos^2x ) - 1 ) ( sqrt(3) + 2 cosx) / (3-4cos^2x) (2sinx + 1)
TrojanPoem
  • TrojanPoem
(4 - 4cos^2x - 1)(sqrt(3) + 2 cosx) / (3-4cos^2x) (2 sinx + 1)
TrojanPoem
  • TrojanPoem
we have 4 - 1 - 4cos^2x which is 3- 4cos^2x ( down)
TrojanPoem
  • TrojanPoem
so we are left with sqrt(3) + 2cosx / 2sinx + 1
TrojanPoem
  • TrojanPoem
sqrt3( + 2 cos 39 . 2 sin30 + 1 = sqrt(3)
phi
  • phi
ok it does work, but it's definitely a case of "lucking out" we get \[ \frac{(4 \sin^2 -1)(\sqrt{3 }+ 2 \cos x)}{(3-4 \cos^2 x)(2 \sin x +1)}\] we can separate that into the limit of the product of two terms. the second term as a limit of sqr(3) \[ \lim_{x \rightarrow \pi/6} \frac{(\sqrt{3 }+ 2 \cos x)}{(2 \sin x +1)} = \sqrt{3}\] the other term needs some work, as it looks like 0/0 but notice \[ 3 - 4 \cos^2 x= 3-4(1-\sin^2 x) \\ = 3 -4 +4 \sin^2 x\\ = 4 \sin^2 x -1 \] and (magically) we have \[ \frac{ 4 \sin^2 x -1}{ 4 \sin^2 x -1} \] which has a limit of 1
TrojanPoem
  • TrojanPoem
xD
TrojanPoem
  • TrojanPoem
but how did the creator of it think of doing that ?
phi
  • phi
In this case it is a lucky trick. You try things (or work with these trig expressions so much you become very good at seeing the way to the end).
TrojanPoem
  • TrojanPoem
I tried every trig formula the end was similar each time a more complicated expression. I am stuck with another limit , may you help ? I will tag you
phi
  • phi
ok

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