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

1. anonymous

k is positive

2. myininaya

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

3. anonymous

yes

4. myininaya

give me a second to think

5. anonymous

wolfram is giving me +/- infinity

6. myininaya

7. anonymous

I'll type out the rest of the word problem

8. 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

9. anonymous

i think its supposed to represent a previous amount

10. myininaya

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

11. myininaya

what is this called?

12. myininaya

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

13. anonymous

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

14. myininaya

and we are assuming maybe n>1?

15. anonymous

That would make sense

16. myininaya

ok i think i got it one sec

17. 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

18. myininaya

as approaches*

19. anonymous

Thank you, that makes sense to me

20. myininaya

cool! :)

21. myininaya