Question

evaluate limit
lim x approaches infinity (1-p/x)^(qx)

Answer 1

as lim x aproaches to infinity (1-p)/x
=e to the power of -pq

Answer 2

\[\lim x->infinity (1-p/x)^{qx}\]

Answer 3

its (1-p)/x^(qx)
= e^-pq

Answer 4

o ok thanks

Answer 5

ah, this is a good one.

case 1: if q = 0 and p isn't 0. simple enough. the limit will simply be 1, because p/x -> 0

case 2: if p = 0 and q isn't 0, simple too. it will be 1 to the power of a huge number (if q is positive). if q is negative, it will be one to the power of a very small number. in either case, it will be 1.

case 3: p is not zero and q is not zero

make the substitution m = qx. as x ->inf, m->oo (since q > 0)

thus the limit becomes (1 - pq/m)^m

= e^(-pq)

Answer 6

and considering when q < 0 leads to a limit of e^(-pq) too

Answer 7

thats a cheap answer

Answer 8

james, how do you know that
the limit (1 - pq/m)^m = e^ -pq

Answer 9

because (1 + x/n)^n, for very large n, converges to e^x

euler's famous limit

Answer 10

ok i guess you can use that result. what if you didnt know it?

Answer 11

prove eulers famous limit then

Answer 12

Good question. I'd use this reasoning:

y = (1 - p/x)^(qx)

ln(y) = qx * ln(1 - p/x)

so ln(y) = ln(1 - p/x) / (1/qx)

Use L'hopital's rule to calculate the limit of ln(y):

= lim x->inf ( -pqx^2/(x^2 - px) )

= lim x->inf ( -pq/(1 - p/x) )

= -pq/(1 - 0)

= -pq

Thus ln(y) converges to -pq

Since ln(y) is continous for all positive y, then y converges to e^(-pq)

Answer 13

nice :)

Answer 14

as for proving l'Hopital's rule, i am unsure. ask l'Hopital :)