calculate-lim-n-0-1-t-n-n-e-3t-dt-

Question Number 94659 by msup by abdo last updated on 20/May/20

c a l c u l a t e {l i m}_{n \to + \infty} \int_{0}^{\infty} {(1 - \frac{t}{n})}^{n} e^{- 3 t} d t

Answered by mathmax by abdo last updated on 20/May/20

let A_{n} = \int_{0}^{\infty} {(1 - \frac{t}{n})}^{n} e^{- 3 t} dt \Rightarrow A_{n} = \int_{R} {(1 - \frac{t}{n})}^{n} e^{- 3 t} χ_{[0, + \infty [} (t) dt

= \int_{R} f_{n} (t) dt we have f_{n} \to^{cs} f = e^{- 4 t} and ∣ f_{n} ∣⩽ e^{- 4 t} \Rightarrow

theorem of convergence dominee give

\lim_{n \to + \infty} A_{n} = \int_{R} {limf}_{n} (t) dt = \int_{0}^{\infty} e^{- 4 t} dt = {[- \frac{1}{4} e^{- 4 t}]}_{0}^{+ \infty} = \frac{1}{4}