Question

a) $\lim_{n \to \infty} \frac{ 2^{n}+1 }{ 2^{n+1}+3 }$
b)$\lim_{n \to \infty} \frac{ 2^{n} + 3^{n+1} }{ 2^{n+1}+ 3^{n} }$
c)$\lim_{n \to \infty} \frac{ 2^{2n+1}+ 3^{n} +1 }{ 3^{n}+ 4^{n} }$
d)$\lim_{n \to \infty} \frac{ 2^{n}+ a^{n} }{ 2^{n}+ 3^{n} }$, a>0

Answer 1

A)Fractia   este   egala   cu(2^n+1)/(2*2^n+3)=dai   2^n factor   comun   foRtaT+
2^n^**(1+1/2^n)/2^n*(2+3/2^n)=1/2   pt   ca  2/n→0   si 3/2^n→0
b)Fractia   se   mai   scrie 
[2^n+3*3^n]/[2*2^n+3^n]=3^n*[(2/3)^n+3]/3^n*(2*(2/3)^n+1=3    pt   ca (2/3)^n→)
pt  ca   e    un   nr   subunitar   ridicart   la   oo

c)fractia   se   mai   scrie[2*2^2n+3^n+1}/(3^n+4^n)=
(2*4^n+3*3^n)/(3^n+4^n)=
4^n*(2+3*(3/4)^n/4^n*[(3/4)^n+1]=2   PT  CA   (3/4)^n→0
D)Dai   pe   a^n   factor   comun   si   obtii
a^n*[(2/a)^n+1]/3^n*[(2/3)^n+1]=
*a/3)^n*[(2/a)^n+1]/3^n*[(2/3)^n+1]=
Discutie
a=3    3^n/3^n=1 L→1   pt  ca  2/3^n→0
a>3  L
(a/3)^n→)   limita   este   0
a>3
(a/n)^n→∞  limita  →∞