## anonymous 3 years ago lim sin^2 x/ 3x^2 x>0 using squeezing theorem .

1. Goten77

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2. anonymous

1/3 ?

3. anonymous

yes ? help me to solve this question ..

4. Goten77

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

5. Goten77

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6. Goten77

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7. Goten77

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8. anonymous

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

9. anonymous

@Goten77 "using squeezing theorem"

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. anonymous

wio ?? can you help me .

15. anonymous

yes I am thinking about it.

16. anonymous

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

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

18. anonymous

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

19. anonymous

Oh, I got an idea now...

20. tkhunny

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

21. anonymous

I don't got it .

22. anonymous

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23. anonymous

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

24. tkhunny

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. anonymous

thanks !

30. anonymous
31. anonymous

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

32. anonymous

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. anonymous

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

34. anonymous

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

35. anonymous

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

36. anonymous

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

37. anonymous

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

38. anonymous

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

39. anonymous

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40. anonymous

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

41. anonymous

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. anonymous

so we have sinx<x<tgx

43. anonymous

huh ?