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yuki

  • 5 years ago

Find the limit as x -> infinity for the following

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  1. Yuki
    • 5 years ago
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    \[\lim_{x \rightarrow \infty} \sin(x/2)/x\]

  2. Yuki
    • 5 years ago
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    \[\lim_{x \rightarrow \infty} x^{1/x}\]

  3. anonymous
    • 5 years ago
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    first ones pinching theorm if I remember correctly lol

  4. anonymous
    • 5 years ago
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    and the second involves taking natural logs

  5. anonymous
    • 5 years ago
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    we know that -1<= sin(x/2) <= 1 divide through by x ( we can do this as x is not zero, it is approaching infinity!) so -1/x <= sin(x/2) / <= 1/x

  6. anonymous
    • 5 years ago
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    now take the limits on the outside , as x-> infinity , they both go to zero , therefore the limit that is pinched between goes to zero

  7. anonymous
    • 5 years ago
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    for second one, let the limit be L L = limit x->infinity ( x^(1/x) ) take ln of both sides so ln(L) = ln [ limit x->infinity x^(1/x) ] but the function "ln" is continuous as n->infinity so we can bring it inside the limit ( there is a continuity theorm that allows us to do this , I dont remember it right now

  8. anonymous
    • 5 years ago
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    hmm, you can use L'hopital's rule for the first one since it's infinity/infinity

  9. anonymous
    • 5 years ago
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    oh nvm , use elec's way ^_^

  10. anonymous
    • 5 years ago
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    so ln(L) = limit (x->infinity ) ln[ x^(1/x) ] RHS = limit (x->infinity ) (1/x) ln(x) [ by a log law ] now this limit is an indeterminate form ( infinity /infinity ) so use La Hopitals rule RHS = limit (x->infinity ) (1/x) / 1 = limit (x->infinity ) (1/x) = 0 but RHS = ln(L) =0 so L= e^0 = 1 The limit is 1. Nice and long questions lol

  11. anonymous
    • 5 years ago
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    the longer the question, the fun it is to solve =P

  12. anonymous
    • 5 years ago
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    Did you get the first one solved?

  13. Yuki
    • 5 years ago
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    I have bad news. the first one the answer is 2

  14. anonymous
    • 5 years ago
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    No, it's 1/2.

  15. anonymous
    • 5 years ago
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    Oh -- hang on...I thought you were taking it to 0. sorry!

  16. anonymous
    • 5 years ago
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    It's 0, not 2, though.

  17. Yuki
    • 5 years ago
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    I got the second one, that was perfect explanation.

  18. anonymous
    • 5 years ago
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    This is a good summary of limit laws: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

  19. Yuki
    • 5 years ago
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    Let me double check the first one. I mean, intuitively sin(x/2) oscilates and x keeps growing so the limit seems to go to 0

  20. anonymous
    • 5 years ago
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    It does go to 0.

  21. Yuki
    • 5 years ago
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    you know what, I believe you guys. the answer said 1/2, but clearly x -> infinity should make it go to 0

  22. Yuki
    • 5 years ago
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    one of those typos maybe.

  23. anonymous
    • 5 years ago
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    you can sub in a really big value for x, say 1 000 000 and see that it works , it not a 100% mathematically correct proof by subing in numbers , but it should help you accept that the limit is zero

  24. anonymous
    • 5 years ago
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    1 000 000 radians off course, need to be in correct mode , which you most likely wont be :P

  25. anonymous
    • 5 years ago
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    You can say,\[\lim_{x \rightarrow \infty}\frac{\sin(x/2)}{x}=\lim_{x \rightarrow \infty} \sin(x/2).\lim_{x \rightarrow \infty}\frac{1}{x}=\lim_{x \rightarrow \infty} \sin(x/2).0=0\]since sin(x/2) is bounded for all x.

  26. Yuki
    • 5 years ago
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    I would like to ask another limit question. how is it shown that \[\lim_{x \rightarrow \infty} (1+1/x)^x = e^x\]

  27. Yuki
    • 5 years ago
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    or was it e?

  28. anonymous
    • 5 years ago
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    similar to the second question I did

  29. Yuki
    • 5 years ago
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    I thought so

  30. anonymous
    • 5 years ago
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    and its just e

  31. anonymous
    • 5 years ago
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    take logs , bring the power down , bring the limit inside the function, use LH's rule etc

  32. Yuki
    • 5 years ago
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    would it be the same for \[\lim_{x \rightarrow \infty} (e^x +1/x )^x\]

  33. Yuki
    • 5 years ago
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    why do I keep doing this? the last exponent is actually "1/x" not "x"

  34. anonymous
    • 5 years ago
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    But e^x is still the same?

  35. anonymous
    • 5 years ago
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    You'd attempt something similar. You have to remove the power.

  36. Yuki
    • 5 years ago
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    yep infinity to the zero power = 1 yaaay ! lol

  37. anonymous
    • 5 years ago
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    You can manipulate it into an indeterminate form and use L'Hopital's.

  38. anonymous
    • 5 years ago
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    \[y=\left( e^x+\frac{1}{x} \right)^{1/x} \rightarrow \log y = \frac{1}{x}\log \left( e^x+\frac{1}{x} \right)=\frac{1/x}{1/\log \left( e^x+\frac{1}{x} \right)}\]

  39. anonymous
    • 5 years ago
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    0/0 is the form as x goes to infinity.

  40. Yuki
    • 5 years ago
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    I was able to do the lim = e thingy. I was always wondering how that works, thanks everyone.

  41. anonymous
    • 5 years ago
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    Or...der...I could have left it as was noting it in the indeterminate form infty/infty.

  42. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow \infty} \log y =\lim_{x \rightarrow \infty} \frac{\log(e^x+1/x)}{x}=\frac{\lim_{x \rightarrow \infty} (\log (e^x+1/x))'}{\lim_{x \rightarrow \infty} (x)'}=\frac{\lim_{x \rightarrow \infty} \frac{e^x-\frac{1}{x^2}}{e^x+\frac{1}{x}}}{\lim_{x \rightarrow \infty} 1}\]

  43. anonymous
    • 5 years ago
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    \[=\lim_{x \rightarrow \infty} \frac{e^x-\frac{1}{x^2}}{e^x+\frac{1}{x}}=\lim_{x \rightarrow \infty} \frac{1-\frac{1}{e^xx^2}}{1+\frac{1}{e^xx}}=1\]

  44. Yuki
    • 5 years ago
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    you are such a math geek lol

  45. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow \infty}\log y = 1 \rightarrow y = e\]

  46. Yuki
    • 5 years ago
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    Thanks everyone, it really helped me a lot. I'm going to take an AP calc diagnostic test soon and that will determine whether I can marry my GF or not. It's on Thursday and I think at this pace I can do a lot better than I thought. Again, thanks people.

  47. anonymous
    • 5 years ago
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    You're welcome, Yuki! Good luck! :D

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