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

1. TrojanPoem

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

2. TrojanPoem

Approaches ( pi/ 6 )

3. phi

I would use L'hopital's rule

4. TrojanPoem

Out of my curriculum.

5. 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

6. TrojanPoem

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

7. TrojanPoem

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

8. TrojanPoem

i have never even seen*

9. phi

I agree it is not an obvious tactic.

10. 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.

11. TrojanPoem

was confused by it*

12. TrojanPoem

By the way , thanks phi.

13. TrojanPoem

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

14. 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)

15. TrojanPoem

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

16. TrojanPoem

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

17. TrojanPoem

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

18. TrojanPoem

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

19. TrojanPoem

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

20. TrojanPoem

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

21. TrojanPoem

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

22. 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

23. TrojanPoem

xD

24. TrojanPoem

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

25. 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).

26. 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

27. phi

ok