Question

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

Answer 1

k is positive

Answer 2

\[\lim_{x \rightarrow 0}\frac{1-x}{x(1+xk)^{n-1}}\]

Answer 3

yes

Answer 4

give me a second to think

Answer 5

wolfram is giving me +/- infinity

Answer 6

what about n

Answer 7

I'll type out the rest of the word problem

Answer 8

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

Answer 9

i think its supposed to represent a previous amount

Answer 10

yeah i don't know about all that stuff
i'm sorry astro

Answer 11

what is this called?

Answer 12

i can try to find someone who knows about whatever this is

Answer 13

no problem- its just studying limits for now based around cell surface receptors

Answer 14

and we are assuming maybe n>1?

Answer 15

That would make sense

Answer 16

ok i think i got it one sec

Answer 17

\[\lim_{x \rightarrow 0^+}\frac{1-x}{x(1+xk)^{n-1}}=\infty \]
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

Answer 18

as approaches*

Answer 19

Thank you, that makes sense to me

Answer 20

cool! :)

Answer 21

i'm glad i could help you with the math part lol