limits calc

Question

the study was made regarding learning theory. students were tested n times and the proportion P of correct answers given by the function.

p=.92/1+e^-.2n

 

what is the limit of P if the students were tested: four times?

an indefinite large number of times?

Answer 1

a/ n=4 so p=.92/(1+e^-.8)=63.5 pct to 1 decimal place

 

b/lim 0.92/(1+e^-.2n) as n to infinity=0.92lim(1-e^-.2n+e^-.3n-.....)=0.92

 

The easy way to see this with reference to the original equation is that the term e^-0.2n=1/e^.2n tends to zero as n tends to infinity because e^.2n tends to infinity.  Note the series in brackets is obtained by long division or the simple formula

 

1/(1-x)=1-x+x^2-x^3+..... if 0<=x<1 which is certainly true for large n since x=e^-.2n