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a fost răspuns

Sa se calculeze limitele urmatoarelor siruri:
a) [tex] \frac{ln(1+ e^{n} )}{n} [/tex]
b)[tex] \frac{ln( n^{3}+n-1 )}{ ln(n^{6} +2 n^{3}+n) } [/tex]
c)[tex] \frac{ln( n^{3}+ e^{2n} )}{ln( n^{6}+ e^{3n} )} [/tex]
d)[tex] (\frac{1}{ e^{n}+ 1}) ^{ \frac{1}{n} } [/tex]
e)[tex] ( 3^{n}+ 5^{n} )^{ \frac{1}{n} } [/tex]

Răspuns :

GeorgeDINFO
1+e^n<2e^n =>
ln(1+e^n)<ln2e^n    n→+∞
lim ln(1+e^n)/n<lm2*e^nImparti  inegalitatea   prin n
limln(1+e^n)/n<lim[ln2+lne^n]/n
limln (1+e^n)/n<lim (ln2+n)/n
limln(1+e^n)/n<lim(ln2/n+n/n
lim(1+1/n)/n<1
ln(1+e^n)>lne^n
 ln(1+e^n)/n>lne^n/n=nlne/n→1
Conform   criteriului  clestelui limita   este=1
e)(3^n+5^n)^1/n=[5^n(3/5)+1)]^1/n=5^n/n*[(3/5)^n+1]^1/n=5  pt ca  (3/5)^n→0

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