calculate-lim-n-1-2i-1-it-n-n-1-it-n-n-

Question Number 36924 by maxmathsup by imad last updated on 07/Jun/18

c a l c u l a t e {l i m}_{n \to + \infty} \frac{1}{2 i} {{(1 + \frac{i t}{n})}^{n} - {(1 - \frac{i t}{n})}^{n})

Commented by math khazana by abdo last updated on 09/Jun/18

w e {h a v e A}_{n} (t) = \frac{1}{2 i} {(1 + \frac{i t}{n})}^{n} - {(1 - \frac{i t}{n})}^{n}

= I m {(1 + \frac{i t}{n})}^{n} b u t ∣ 1 + \frac{i t}{n} ∣= \sqrt{1 + \frac{t^{2}}{n^{2}}}

= \frac{\sqrt{t^{2} + n^{2}}}{n} \Rightarrow 1 + \frac{i t}{n} = \frac{\sqrt{t^{2} + n^{2}}}{n} {\frac{n}{\sqrt{t^{2} + n^{2}}} + i \frac{t}{\sqrt{t^{2} + n^{2}}}}

= r e^{i θ} \Rightarrow r = \frac{\sqrt{t^{2} + n^{2}}}{n} a n d c o s θ = \frac{n}{\sqrt{t^{2} + n^{2}}}

s i n θ = \frac{t}{\sqrt{t^{2} + n^{2}}} \Rightarrow t a n (θ) = \frac{t}{n} \Rightarrow θ = a r t a n (\frac{t}{n})

\Rightarrow (1 + \frac{i t}{n}) = r e^{i a r c t a n (\frac{t}{n})} \Rightarrow A_{n} (t) = r^{n} e^{i n a r c t a n (\frac{t}{n})}

r = \sqrt{1 + \frac{t^{2}}{n^{2}}} \sim 1 + \frac{t^{2}}{2 n^{2}} \Rightarrow r^{n} \sim {(1 + \frac{t^{2}}{2 n^{2}})}^{n}

\sim 1 + \frac{t^{2}}{2 n} \to 1 (n \to + \infty)

a r c t a n (\frac{t}{n}) \sim \frac{t}{n} (n \to + \infty) a n d n a t c t a n (\frac{t}{n}) \sim t

s o {l i m}_{n \to + \infty} A_{n} (t) = e^{i t} = c o s t + i s i n t .