# lim V, m-->∞ =?

V = [(mg)/c][1-e^((-ct)/m)]

lim V m-->∞ =?

Please help me *understand* the basic method used to solve this. I tried ln on both sides and also a method with e, but fell short.

### 4 Answers

- anil bakshiLv 71 decade agoBest Answer
lim V, m-->∞ =?

V = [mg / c] [1 - e^(-ct)/m)]

= g* [1 - e^(-ct)/m)] / [c / m] (0/0) form ------(1)

let c / m = k

V = g* [1 - e^-kt] / [k]

--------------------------------------------------

e^-kt = 1 - kt + (kt)^2 /2! - (kt)^3 / 3!+......

[1 - e^-kt] = kt - (kt)^2 /2! + (kt)^3 / 3!+......

divide both sides by k

{[1 - e^-kt] / k } = t - k (t)^2 /2! + k^2(t)^3 / 3!+......

----------------------------------------------------

put in (1)

V = g*{t - k (t)^2 /2! + k^2(t)^3 / 3!+...} (use >> c / m = k)

m-->∞ , k >>0 apply limit

V = g*{ t - o (t)^2 /2! + o (t)^3 / 3!+...}

V = g * t

lim V, m-->∞ =?

V = [(mg)/c][1-e^((-ct)/m)] = g * t answer

- hustolemynameLv 61 decade ago
V = (mg/c) [ ct/m +O(1/m^2) ] = gt + O(1/m) -->gt

where O(m^r) means something/anything of the order of m^r

- Anonymous1 decade ago
This has indeterminate form ∞*0. Rewrite the expression as the second factor divided by c/(mg), which is the reciprocal of the first factor, and now you have indeterminate form 0/0. Use L'Hopital's rule on that.

- Anonymous1 decade ago
it's not easy !