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anonymous

  • 4 years ago

lim x approaches 0 for f(x)=(1-x)/(x(1+xk)^(n-1))

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  1. anonymous
    • 4 years ago
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    k is positive

  2. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}\frac{1-x}{x(1+xk)^{n-1}}\]

  3. anonymous
    • 4 years ago
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    yes

  4. myininaya
    • 4 years ago
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    give me a second to think

  5. anonymous
    • 4 years ago
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    wolfram is giving me +/- infinity

  6. myininaya
    • 4 years ago
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    what about n

  7. anonymous
    • 4 years ago
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    I'll type out the rest of the word problem

  8. anonymous
    • 4 years ago
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    In an article on the local clustering of cell surface receptors, the researcher analyzed the limits of the function: (The function I put) X is the concentration of free receptors, n is the number of functional groups of cells, and k is a positive constant The function is also equal to C which is the concentration of free ligand in the medium

  9. anonymous
    • 4 years ago
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    i think its supposed to represent a previous amount

  10. myininaya
    • 4 years ago
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    yeah i don't know about all that stuff i'm sorry astro

  11. myininaya
    • 4 years ago
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    what is this called?

  12. myininaya
    • 4 years ago
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    i can try to find someone who knows about whatever this is

  13. anonymous
    • 4 years ago
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    no problem- its just studying limits for now based around cell surface receptors

  14. myininaya
    • 4 years ago
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    and we are assuming maybe n>1?

  15. anonymous
    • 4 years ago
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    That would make sense

  16. myininaya
    • 4 years ago
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    ok i think i got it one sec

  17. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0^+}\frac{1-x}{x(1+xk)^{n-1}}=\infty \] since 1-x is negative for any values approaching 0 from the right but both the factors of the bottom are positive since k>0 we know nothing can be done to force f to be continuous at x=0 so we know from the right it approaches +infinity and from the left it approaches negative infinity since 1-x is positive at x approaches 0 from the left

  18. myininaya
    • 4 years ago
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    as approaches*

  19. anonymous
    • 4 years ago
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    Thank you, that makes sense to me

  20. myininaya
    • 4 years ago
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    cool! :)

  21. myininaya
    • 4 years ago
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    i'm glad i could help you with the math part lol

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