## Babynini one year ago Use squeeze theorem to show that lim sqroot (x^3+x^2)sin(pi/x)=0 x-> 0 illustrate by graphing the functions f,g,h on the same screen.

1. Babynini

|dw:1443761963714:dw|

2. Babynini

@zepdrix

3. Babynini

@Astrophysics

4. zepdrix

Which piece is causing trouble? $$\large\rm \sqrt{x^3+x^2}$$ or the $$\large\rm \sin\left(\frac{\pi}{x}\right)$$? :)

5. Babynini

I'm not sure where to start xD

6. zepdrix

OpenStudy lag :O( lameeeee

7. Babynini

hmm?

8. anonymous

As we know, sin of any number will always be between -1 and 1, inclusive.

9. anonymous

$-1\le \sin \frac{ \pi }{ x }\le1$

10. anonymous

now we multiply whole thing by $\sqrt{x ^{3}+x ^{2}}$

11. Babynini

multiply the whole sin(pi/x)?

12. zepdrix

Plug in x=0, which part of the function is causing a problem?

13. anonymous

yes like this

14. anonymous

its a squezz theorm @zepdrix

15. anonymous

ok i ll do it step by step u ll understand let me finish

16. Babynini

(@zepdrix the sin part?) okay @ayeshaafzal221 :)

17. zepdrix

..

18. anonymous

$-(\sqrt{x ^{3}+x ^{2}})\le \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \sqrt{x ^{3}+x ^{2}}$

19. zepdrix

Yes, so start by putting bounds on the bad part. Sine function is bound by -1 and 1, yes? :)

20. Babynini

Yeah

21. anonymous

With an inequality like this, you can add "lim x->0" to all 3 parts and the expression will still be valid.

22. zepdrix

$\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le 1$

23. Babynini

Okay, @zepdrix next what? :)

24. anonymous

$\lim_{x \rightarrow 0}-(\sqrt{x ^{3}+x ^{2}})\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}}$

25. anonymous

now you solve left hand side and then right hand side

26. zepdrix

:p

27. anonymous

so left hand side is $\lim_{x \rightarrow 0} -(\sqrt{0^{3} +0^{2}} =-\sqrt{0}=0$

28. anonymous

i am sub x=0

29. anonymous

now right hand side $\lim_{x \rightarrow 0}\sqrt{0^{3}+0^{2}}=0$

30. anonymous

thus the limit of the original expression is less than and equal to 0, and greater and equal to 0, then its limit must be 0.

31. Babynini

ooo so, to answer the question, to I have to write out that proof?

32. anonymous

if u didnt understood any step let me know and i know what you confuse about whenever there is squeez theorm there are restriction that u must know before hand for cos , tan and sin , u can google em

33. Babynini

The question says to "use notation of the squeeze theorem"

34. anonymous

yes :) thats what squeex theorm is

35. Babynini

Okay, because I don't remember multiplying things in doing limits in class o.o

36. anonymous

ur question clearly says use squeez theorm to show that expression =0

37. Babynini

Yeah. Usually we approach the limit from the left and right and see if they match up o.0 or calculate it as it approaches 0

38. anonymous

yup thats what i did

39. anonymous

40. anonymous

see these domain and range u must know by heart its really useful

41. anonymous

in ur case just remember cos , sin and tan

42. Babynini

ahh ok

43. Babynini

I am beginning to understand.

44. anonymous

good to know just tag if u have any other problem :)

45. Babynini

ok thanks!!

46. Babynini

i) -1 </= sin (pi/x) </= 1 ii) $-\sqrt{x^3+x^2}\le \sqrt{x^3+x^2}\sin(\frac{ \pi }{ x })\le \sqrt{x^3+x^2}$ iii) we know that $\lim \sqrt{x^3+^2} = 0 \space and \space \lim(-\sqrt{x^3+x^2}=0$ Taking f(x)= -sq( x^3+x^2), g(x)= sq(x^3+x^2)sin(pi/x) and, h(x) = sq(x^3+x^2), in squeeze theorem, we conclude: lim sq(x^3+x^2)sin(pi/)=0 (as approaches 0)

47. Babynini

*sin(pi/x)

48. Babynini

@ayeshaafzal221 does that all look correct?

49. anonymous

yes @Babynini well done u got a hang of it :)

50. Babynini

Thank you! :)