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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.

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  1. Babynini
    • one year ago
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    |dw:1443761963714:dw|

  2. Babynini
    • one year ago
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    @zepdrix

  3. Babynini
    • one year ago
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    @Astrophysics

  4. zepdrix
    • one year ago
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    Which piece is causing trouble? \(\large\rm \sqrt{x^3+x^2}\) or the \(\large\rm \sin\left(\frac{\pi}{x}\right)\)? :)

  5. Babynini
    • one year ago
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    I'm not sure where to start xD

  6. zepdrix
    • one year ago
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    OpenStudy lag :O( lameeeee

  7. Babynini
    • one year ago
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    hmm?

  8. anonymous
    • one year ago
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    As we know, sin of any number will always be between -1 and 1, inclusive.

  9. anonymous
    • one year ago
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    \[-1\le \sin \frac{ \pi }{ x }\le1\]

  10. anonymous
    • one year ago
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    now we multiply whole thing by \[\sqrt{x ^{3}+x ^{2}}\]

  11. Babynini
    • one year ago
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    multiply the whole sin(pi/x)?

  12. zepdrix
    • one year ago
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    Plug in x=0, which part of the function is causing a problem?

  13. anonymous
    • one year ago
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    yes like this

  14. anonymous
    • one year ago
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    its a squezz theorm @zepdrix

  15. anonymous
    • one year ago
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    ok i ll do it step by step u ll understand let me finish

  16. Babynini
    • one year ago
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    (@zepdrix the sin part?) okay @ayeshaafzal221 :)

  17. zepdrix
    • one year ago
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    ..

  18. anonymous
    • one year ago
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    \[-(\sqrt{x ^{3}+x ^{2}})\le \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \sqrt{x ^{3}+x ^{2}}\]

  19. zepdrix
    • one year ago
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    Yes, so start by putting bounds on the bad part. Sine function is bound by -1 and 1, yes? :)

  20. Babynini
    • one year ago
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    Yeah

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

  22. zepdrix
    • one year ago
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    \[\large\rm -1\le \sin\left(\frac{\pi}{x}\right)\le 1\]

  23. Babynini
    • one year ago
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    Okay, @zepdrix next what? :)

  24. anonymous
    • one year ago
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    \[\lim_{x \rightarrow 0}-(\sqrt{x ^{3}+x ^{2}})\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}} \times \sin \frac{ \pi }{ x }\le \lim_{x \rightarrow 0} \sqrt{x ^{3}+x ^{2}}\]

  25. anonymous
    • one year ago
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    now you solve left hand side and then right hand side

  26. zepdrix
    • one year ago
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    :p

  27. anonymous
    • one year ago
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    so left hand side is \[\lim_{x \rightarrow 0} -(\sqrt{0^{3} +0^{2}} =-\sqrt{0}=0\]

  28. anonymous
    • one year ago
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    i am sub x=0

  29. anonymous
    • one year ago
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    now right hand side \[\lim_{x \rightarrow 0}\sqrt{0^{3}+0^{2}}=0\]

  30. anonymous
    • one year ago
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    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
    • one year ago
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    ooo so, to answer the question, to I have to write out that proof?

  32. anonymous
    • one year ago
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    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
    • one year ago
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    The question says to "use notation of the squeeze theorem"

  34. anonymous
    • one year ago
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    yes :) thats what squeex theorm is

  35. Babynini
    • one year ago
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    Okay, because I don't remember multiplying things in doing limits in class o.o

  36. anonymous
    • one year ago
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    ur question clearly says use squeez theorm to show that expression =0

  37. Babynini
    • one year ago
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    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
    • one year ago
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    yup thats what i did

  39. anonymous
    • one year ago
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    1 Attachment
  40. anonymous
    • one year ago
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    see these domain and range u must know by heart its really useful

  41. anonymous
    • one year ago
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    in ur case just remember cos , sin and tan

  42. Babynini
    • one year ago
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    ahh ok

  43. Babynini
    • one year ago
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    I am beginning to understand.

  44. anonymous
    • one year ago
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    good to know just tag if u have any other problem :)

  45. Babynini
    • one year ago
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    ok thanks!!

  46. Babynini
    • one year ago
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    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
    • one year ago
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    *sin(pi/x)

  48. Babynini
    • one year ago
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    @ayeshaafzal221 does that all look correct?

  49. anonymous
    • one year ago
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    yes @Babynini well done u got a hang of it :)

  50. Babynini
    • one year ago
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    Thank you! :)

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