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TrojanPoem
 one year ago
Find the limit of : ( No L'hopital's rule)
TrojanPoem
 one year ago
Find the limit of : ( No L'hopital's rule)

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TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \frac{ \pi }{ 6 }} \frac{ 2\sin x 1 }{ \sqrt{3}  2\cos x }\]

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Approaches ( pi/ 6 )

phi
 one year ago
Best ResponseYou've already chosen the best response.1I would use L'hopital's rule

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0Out of my curriculum.

phi
 one year ago
Best ResponseYou've already chosen the best response.1one trick that often works is to multiply by the conjugate of the denominator the idea is (a+b)(ab) = a^2  b^2

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0But what made us think of doing so. To me It's clear when there is square roots.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0I even never so example when they multiply by the conjugate with sin , cos , tan

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0i have never even seen*

phi
 one year ago
Best ResponseYou've already chosen the best response.1I agree it is not an obvious tactic.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0After 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
 one year ago
Best ResponseYou've already chosen the best response.0was confused by it*

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0By the way , thanks phi.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0If you ever though of trying to it and found alternative solution , Pm me !

phi
 one year ago
Best ResponseYou've already chosen the best response.1multiplying 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
 one year ago
Best ResponseYou've already chosen the best response.0Give me a second , I will see the solution written in my notebook.

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(4sin^2x  1) (sqrt(3) + 2cosx) / (3  4 cos ^2 x) (2 sinx +1 )

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(4( 1 cos^2x )  1 ) ( sqrt(3) + 2 cosx) / (34cos^2x) (2sinx + 1)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0(4  4cos^2x  1)(sqrt(3) + 2 cosx) / (34cos^2x) (2 sinx + 1)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0we have 4  1  4cos^2x which is 3 4cos^2x ( down)

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0so we are left with sqrt(3) + 2cosx / 2sinx + 1

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0sqrt3( + 2 cos 39 . 2 sin30 + 1 = sqrt(3)

phi
 one year ago
Best ResponseYou've already chosen the best response.1ok it does work, but it's definitely a case of "lucking out" we get \[ \frac{(4 \sin^2 1)(\sqrt{3 }+ 2 \cos x)}{(34 \cos^2 x)(2 \sin x +1)}\] 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 \[ 3  4 \cos^2 x= 34(1\sin^2 x) \\ = 3 4 +4 \sin^2 x\\ = 4 \sin^2 x 1 \] and (magically) we have \[ \frac{ 4 \sin^2 x 1}{ 4 \sin^2 x 1} \] which has a limit of 1

TrojanPoem
 one year ago
Best ResponseYou've already chosen the best response.0but how did the creator of it think of doing that ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1In 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
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