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anonymous
 4 years ago
lim x approaches 0 for f(x)=(1x)/(x(1+xk)^(n1))
anonymous
 4 years ago
lim x approaches 0 for f(x)=(1x)/(x(1+xk)^(n1))

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myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0}\frac{1x}{x(1+xk)^{n1}}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2give me a second to think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wolfram is giving me +/ infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'll type out the rest of the word problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In 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
 4 years ago
Best ResponseYou've already chosen the best response.0i think its supposed to represent a previous amount

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2yeah i don't know about all that stuff i'm sorry astro

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2i can try to find someone who knows about whatever this is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem its just studying limits for now based around cell surface receptors

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2and we are assuming maybe n>1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That would make sense

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2ok i think i got it one sec

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow 0^+}\frac{1x}{x(1+xk)^{n1}}=\infty \] since 1x 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 1x is positive at x approaches 0 from the left

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you, that makes sense to me

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.2i'm glad i could help you with the math part lol
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