anonymous
  • anonymous
lim x approaches 0 for f(x)=(1-x)/(x(1+xk)^(n-1))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
k is positive
myininaya
  • myininaya
\[\lim_{x \rightarrow 0}\frac{1-x}{x(1+xk)^{n-1}}\]
anonymous
  • anonymous
yes

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myininaya
  • myininaya
give me a second to think
anonymous
  • anonymous
wolfram is giving me +/- infinity
myininaya
  • myininaya
what about n
anonymous
  • anonymous
I'll type out the rest of the word problem
anonymous
  • anonymous
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
anonymous
  • anonymous
i think its supposed to represent a previous amount
myininaya
  • myininaya
yeah i don't know about all that stuff i'm sorry astro
myininaya
  • myininaya
what is this called?
myininaya
  • myininaya
i can try to find someone who knows about whatever this is
anonymous
  • anonymous
no problem- its just studying limits for now based around cell surface receptors
myininaya
  • myininaya
and we are assuming maybe n>1?
anonymous
  • anonymous
That would make sense
myininaya
  • myininaya
ok i think i got it one sec
myininaya
  • myininaya
\[\lim_{x \rightarrow 0^+}\frac{1-x}{x(1+xk)^{n-1}}=\infty \] 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
myininaya
  • myininaya
as approaches*
anonymous
  • anonymous
Thank you, that makes sense to me
myininaya
  • myininaya
cool! :)
myininaya
  • myininaya
i'm glad i could help you with the math part lol

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