## nurnabilashafiqah Group Title lim sin^2 x/ 3x^2 x>0 using squeezing theorem . one year ago one year ago

1. Goten77 Group Title

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2. akim0001 Group Title

1/3 ?

3. nurnabilashafiqah Group Title

yes ? help me to solve this question ..

4. Goten77 Group Title

well u can use locotoles rule |dw:1356631309095:dw|

5. Goten77 Group Title

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6. Goten77 Group Title

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7. Goten77 Group Title

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8. nurnabilashafiqah Group Title

the 2nd last solution .. it's blank ?? the last equation ..how it can that ?

9. wio Group Title

@Goten77 "using squeezing theorem"

10. wio Group Title

How about?$\large \frac{-1}{ 3x^2} \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2}$

11. wio Group Title

Hmm or maybe even $\large 0 \leq \frac{\sin^2 (x)}{ 3x^2} \leq \frac{1}{ 3x^2}$

12. wio Group Title

Hmm, no, these won't work because they're not defined at 0...

13. nurnabilashafiqah Group Title

sin x . cos x become -sin^2 . cos^2 ??

14. nurnabilashafiqah Group Title

wio ?? can you help me .

15. wio Group Title

yes I am thinking about it.

16. wio Group Title

You need to find two functions... $$g(x)$$ and $$f(x)$$. The following must be true: $\lim_{x \to 0}g(x) = L \\ \lim_{x \to 0}f(x) = L \\ g(x) \leq \frac{\sin^2(x)}{3x^2} \leq f(x)$

17. tkhunny Group Title

Have you considered $$\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1$$?

18. nurnabilashafiqah Group Title

tkhunny : yes . I know about it . first rule in squeezing theorem, right ?

19. wio Group Title

Oh, I got an idea now...

20. tkhunny Group Title

Using that, we can find the limit directly. How do you suppose we inflict the Squeeze Theorem upon it?

21. nurnabilashafiqah Group Title

I don't got it .

22. RaphaelFilgueiras Group Title

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23. nurnabilashafiqah Group Title

ouh . okay . now I get it ! thank you <3

24. tkhunny Group Title

How does $$\left(\dfrac{\sin(x)}{x}\right)^{2}$$ compare to $$\dfrac{sin(x)}{x}$$ for $$x \in (0,1)$$

25. nurnabilashafiqah Group Title

just separate it, right ? @RaphaelFilgueiras, @wio, @tkhunny

26. wio Group Title

$$x\in [-1, 1]$$

27. wio Group Title

$$\epsilon > 0,\quad x\in [0-\epsilon, 0 + \epsilon]$$

28. RaphaelFilgueiras Group Title

x must be in radians,this is the only restriction to aply the fundamental trygonometric limit

29. nurnabilashafiqah Group Title

thanks !

30. RaphaelFilgueiras Group Title
31. wio Group Title

@nurnabilashafiqah That's not using the squeeze theorem, so it's not what the question is asking of you.

32. wio Group Title

You need to find two functions that are equal to 1/3 at 0, one function must be greater than your function, the other must be less than your function

33. wio Group Title

Use something like the fact that $\sin^2(x) \leq x^2$

34. wio Group Title

That can get you an upperbound. Then you need a lower bound.

35. RaphaelFilgueiras Group Title

@wio is rigth i did not see the part of squeezing

36. sirm3d Group Title

$\large \cos x < \frac{\sin x}{x} < \frac{1}{\cos x}$

37. RaphaelFilgueiras Group Title

but if you are in radian you will have 0 < sinx²<x²

38. nurnabilashafiqah Group Title

squeezing theorem have both less than and greater than ? sorry, I'm new about this topic . please explain to me briefly .

39. RaphaelFilgueiras Group Title

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40. RaphaelFilgueiras Group Title

that arc is x because in radian system we use 1 as radius

41. wio Group Title

Okay so I'm thinking the best to use is: $\frac{\cos^2(x) }{3} \leq \frac{\sin^2(x)}{3x^2} \leq \frac{x^2}{3x^2}$

42. RaphaelFilgueiras Group Title

so we have sinx<x<tgx

43. nurnabilashafiqah Group Title

huh ?