calculate-lim-n-n-2-e-sin-pi-n-2-cos-pi-n-

Question Number 73334 by mathmax by abdo last updated on 10/Nov/19

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

Answered by Smail last updated on 10/Nov/19

s i n (\frac{π}{n^{2}}) \underset{\infty}{\sim} \frac{π}{n^{2}} a n d e^{1 / n} \underset{\infty}{\sim} 1 + \frac{1}{n}

e^{s i n (π / n^{2})} \underset{\infty}{\sim} 1 + \frac{π}{n^{2}} a n d c o s (\frac{π}{n}) \underset{\infty}{\sim} 1 - \frac{π^{2}}{2 n^{2}}

e^{s i n (π / n^{2})} - c o s (π / n) \underset{\infty}{\sim} 1 + \frac{π}{n^{2}} - 1 + \frac{π^{2}}{2 n^{2}}

\underset{\infty}{\sim} \frac{2 π + π^{2}}{2 n^{2}}

s o {\underset{n \to \infty}{l i m n}}^{2} (e^{s i n (π / n^{2})} - c o s (π / n)) = {\underset{n \to \infty}{l i m n}}^{2} \times (\frac{2 π + π^{2}}{2 n^{2}})

= \frac{2 π + π^{2}}{2}

Commented by mathmax by abdo last updated on 10/Nov/19

a r e y o u n o w i n u s a s i r s m a i l . .

Commented by Smail last updated on 10/Nov/19

Y e s, I c u r r e n t l y l i v e i n t h e U S

Facebook Tweet Pin