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anonymous

  • 4 years ago

Another one of those limit problems, if it's approaching from the left or right it gives a different result: Use the definitions of the hyperbolic functions to find the following limit: cosh (4) And when I do lim x->0 (e^x + e^-x)/ (e^x-e^-x) I get 2 results

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  1. anonymous
    • 4 years ago
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    Sorry its cothx

  2. anonymous
    • 4 years ago
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  3. anonymous
    • 4 years ago
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    Interesting. You are correct, the limits from the left and right are not the same.

  4. anonymous
    • 4 years ago
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    Hmm, I don't really know how to answer this one, if you were to guess, have any ideas on what would be my best bet?

  5. anonymous
    • 4 years ago
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    myininaya, you were typing for a while, did ya have any ideas?

  6. anonymous
    • 4 years ago
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    The limit does not exist, so I don't know what to tell ya :/

  7. Mertsj
    • 4 years ago
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    I thought that the rule is that if the limit from the right is different from the limit from the left, then the limit does not exist. Am I wrong?

  8. anonymous
    • 4 years ago
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    No, that's correct.

  9. anonymous
    • 4 years ago
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    So you think I should go for 0?

  10. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0^+}\frac{e^x+e^{-x}}{e^x-e^{-x}} \cdot \frac{e^x}{e^x}\] \[\lim_{x \rightarrow 0^+}\frac{e^{2x}+1}{e^{2x}-1}\] \[\text{ let } u=e^{2x} ; \text{ as } x->0^+ , u->1\] \[\lim_{u \rightarrow 1}\frac{u+1}{u-1}\] We have vertical asymptote at u=1

  11. myininaya
    • 4 years ago
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    lets make that as \[u->1^+\]

  12. myininaya
    • 4 years ago
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    so we know that as u->1^+ then (u+1)/(u-1)->infinty

  13. myininaya
    • 4 years ago
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    now if we look at as u->1^- then (u+1)/(u-1)-> -infinity

  14. anonymous
    • 4 years ago
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    Yeah, and neither one of those are an option on the question

  15. anonymous
    • 4 years ago
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    They both are but the point is the limit doesn't exist.

  16. anonymous
    • 4 years ago
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    So do you have any ideas on what I should do?

  17. myininaya
    • 4 years ago
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    the limit does not exist

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