anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Sorry its cothx
anonymous
  • anonymous
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anonymous
  • anonymous
Interesting. You are correct, the limits from the left and right are not the same.

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anonymous
  • anonymous
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?
anonymous
  • anonymous
myininaya, you were typing for a while, did ya have any ideas?
anonymous
  • anonymous
The limit does not exist, so I don't know what to tell ya :/
Mertsj
  • Mertsj
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?
anonymous
  • anonymous
No, that's correct.
anonymous
  • anonymous
So you think I should go for 0?
myininaya
  • myininaya
\[\lim_{x \rightarrow 0^+}\frac{e^x+e^{-x}}{e^x-e^{-x}} \cdot \frac{e^x}{e^x}\] \[\lim_{x \rightarrow 0^+}\frac{e^{2x}+1}{e^{2x}-1}\] \[\text{ let } u=e^{2x} ; \text{ as } x->0^+ , u->1\] \[\lim_{u \rightarrow 1}\frac{u+1}{u-1}\] We have vertical asymptote at u=1
myininaya
  • myininaya
lets make that as \[u->1^+\]
myininaya
  • myininaya
so we know that as u->1^+ then (u+1)/(u-1)->infinty
myininaya
  • myininaya
now if we look at as u->1^- then (u+1)/(u-1)-> -infinity
anonymous
  • anonymous
Yeah, and neither one of those are an option on the question
anonymous
  • anonymous
They both are but the point is the limit doesn't exist.
anonymous
  • anonymous
So do you have any ideas on what I should do?
myininaya
  • myininaya
the limit does not exist

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