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Jusaquikie

  • 2 years ago

lim 8e^(TanX) x → (π/2)+

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  1. Jusaquikie
    • 2 years ago
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    not sure where to go to with this

  2. TuringTest
    • 2 years ago
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    since\[\Large\lim_{x\to a}e^{f(x)}=e^{\lim_{x\to a}f(x)}\]really all you need to know is the limit\[\Large\lim_{x\to\pi/2^+}\tan x\]

  3. TuringTest
    • 2 years ago
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    what is\[\lim_{x\to\pi/2}\tan x\]

  4. TuringTest
    • 2 years ago
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    ?

  5. Jusaquikie
    • 2 years ago
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    +infinity?

  6. Jusaquikie
    • 2 years ago
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    .027

  7. TuringTest
    • 2 years ago
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    actually it depends on the left and right hand approach from the left (x<pi/2) cos x>0 from the right (x>pi/2) cos x<0

  8. TuringTest
    • 2 years ago
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    I have no idea where you got that number from....

  9. Jusaquikie
    • 2 years ago
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    tan(pi/2) lol

  10. TuringTest
    • 2 years ago
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    I'm gonna take a wild guess and say you left your calculator in degree mode

  11. Jusaquikie
    • 2 years ago
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    yes

  12. TuringTest
    • 2 years ago
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    \[\tan x=\frac{\sin x}{\cos x}\]so\[\tan (\pi/2)=\frac{\sin (\pi/2)}{\cos (\pi/2)}=?\]

  13. Jusaquikie
    • 2 years ago
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    1/0= undefined

  14. TuringTest
    • 2 years ago
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    right, now what about coming from the right, x>pi/2 will cosince be positive or negative approaching pi/2 from the right?

  15. TuringTest
    • 2 years ago
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    cosine*

  16. Jusaquikie
    • 2 years ago
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    positive and increasing to 1?

  17. TuringTest
    • 2 years ago
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    no, what is the cosine of pi/2 ?

  18. Jusaquikie
    • 2 years ago
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    0 sorry was thinking sin

  19. Jusaquikie
    • 2 years ago
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    so negative

  20. TuringTest
    • 2 years ago
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    correct, so as \(x\to\pi/2^+\) we have that \(\cos x\to0\), which means that\[\lim_{x\to\pi/2^+}\tan x=?\]

  21. Jusaquikie
    • 2 years ago
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    0

  22. TuringTest
    • 2 years ago
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    no, think in terms of sin and cos

  23. Jusaquikie
    • 2 years ago
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    anything with Euler's number just confuses me, i'm not sure how to treat it

  24. Jusaquikie
    • 2 years ago
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    2pi

  25. TuringTest
    • 2 years ago
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    ignore Euler's number, it could be any exponential base, the answer would be the same...\[\lim_{x\to\pi/2^+}\frac{\sin x}{\cos x}=?\]what is sine approaching? what is cos approaching?

  26. TuringTest
    • 2 years ago
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    what is sine approaching? what is cos approaching?

  27. Jusaquikie
    • 2 years ago
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    1,0

  28. TuringTest
    • 2 years ago
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    and is that zero being approached from the negative or positive side?

  29. Jusaquikie
    • 2 years ago
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    negative?

  30. TuringTest
    • 2 years ago
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    correct, so considering that\[\lim_{x\to\pi/2^+}\tan x=\lim_{x\to\pi/2^+}\frac{\sin x}{\cos x}\to\frac10\]and that that zero is being approached from the negative side, what is the limit?

  31. Jusaquikie
    • 2 years ago
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    i can't visualize it and i'm not sure how to graph it in my calculator so i'm trying to relate it to the unit circle

  32. Jusaquikie
    • 2 years ago
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    -infinity?

  33. Jusaquikie
    • 2 years ago
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    no + infinity

  34. TuringTest
    • 2 years ago
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    |dw:1347837808350:dw|you were right the first time, -infty

  35. TuringTest
    • 2 years ago
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    positive number (sine) being divided by a small negative number (cosine) is a large negative number

  36. TuringTest
    • 2 years ago
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    \[\frac1{-0.1}=-10\]\[\frac1{-0.01}=-100\]\[\frac1{-0.001}=-1000\]etc., so as cos x goes to zero from x>pi/2 we approach \(-\infty\)

  37. TuringTest
    • 2 years ago
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    hence\[\lim_{x\to\pi/2^+}\tan x=-\infty\]

  38. Jusaquikie
    • 2 years ago
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    ok

  39. TuringTest
    • 2 years ago
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    so then what is\[\lim_{x\to\pi/2^+}8e^{\tan x}\]

  40. Jusaquikie
    • 2 years ago
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    zero?

  41. TuringTest
    • 2 years ago
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    correct :)

  42. Jusaquikie
    • 2 years ago
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    thanks for the long journey, i'm just really tired and burnt out right now, sorry you had to work so hard on this one

  43. TuringTest
    • 2 years ago
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    it's fine, much better than just pumping out an answer you won't comprehend hopefully you learned something is the idea ;)

  44. Jusaquikie
    • 2 years ago
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    yes thanks

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