Find the limit of : ( No L'hopital's rule)

- TrojanPoem

Find the limit of : ( No L'hopital's rule)

- schrodinger

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- TrojanPoem

\[\lim_{x \rightarrow \frac{ \pi }{ 6 }} \frac{ 2\sin x -1 }{ \sqrt{3} - 2\cos x }\]

- TrojanPoem

Approaches ( pi/ 6 )

- phi

I would use L'hopital's rule

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## More answers

- TrojanPoem

Out of my curriculum.

- 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

But what made us think of doing so. To me It's clear when there is square roots.

- TrojanPoem

I even never so example when they multiply by the conjugate with sin , cos , tan

- TrojanPoem

i have never even seen*

- phi

I agree it is not an obvious tactic.

- 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

was confused by it*

- TrojanPoem

By the way , thanks phi.

- TrojanPoem

If you ever though of trying to it and found alternative solution , Pm me !

- 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

Give me a second , I will see the solution written in my notebook.

- TrojanPoem

(4sin^2x - 1) (sqrt(3) + 2cosx) / (3 - 4 cos ^2 x) (2 sinx +1 )

- TrojanPoem

(4( 1- cos^2x ) - 1 ) ( sqrt(3) + 2 cosx) / (3-4cos^2x) (2sinx + 1)

- TrojanPoem

(4 - 4cos^2x - 1)(sqrt(3) + 2 cosx) / (3-4cos^2x) (2 sinx + 1)

- TrojanPoem

we have 4 - 1 - 4cos^2x which is 3- 4cos^2x ( down)

- TrojanPoem

so we are left with sqrt(3) + 2cosx / 2sinx + 1

- TrojanPoem

sqrt3( + 2 cos 39 . 2 sin30 + 1 = sqrt(3)

- 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

xD

- TrojanPoem

but how did the creator of it think of doing that ?

- 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

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

ok

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