Answer :
Answer:
Step-by-step explanation:
If a sequence c1,c2,c3,...has limit K then the sequence ec1,ec2,ec3,...has limit e^K. Use this fact together with l'Hopital's rule to compute the limit of the sequence given by
bn=(n)^(5.6/n).
a)
[tex]L = \lim_{n \to \infty} b_n \\\\\\L= \lim_{n \to \infty} n^{\frac{5.6}{n} }[/tex]
Log on both sides
[tex]In (L) = \lim_{n \to \infty} In (n)^{\frac{5.6}{n} }\\\\= \lim_{n \to \infty} \frac{5.6}{n} In(n)[/tex]
[tex]=5.6 \lim_{n \to \infty} \frac{d}{dn} In(n)/\frac{d}{dn} (n)\\\\=5.6 \lim_{n \to \infty} \frac{1}{n} /1 \\\\=5.6 \lim_{n \to \infty} \frac{1}{n} \\\\=5.6 \times 0\\\\In(L) =0\\\\L=e^0\\\\L=1[/tex]
[tex]\therefore \lim_{n \to \infty} (n)^{\frac{5.6}{n} =1[/tex]
The limit value of given sequece is 1.
To understand more, check below explanation.
Limit of function:
The given sequence is,
[tex]b_{n}=n^{5.6/n}[/tex]
We have to find limit of above sequence.
[tex]L=\lim_{n \to \infty} b_n \\\\L=\lim_{n \to \infty}n^{5.6/n} \\\\ln(L)=\lim_{n \to \infty}\frac{5.6}{n}ln(n) \\\\ln(L)=5.6\lim_{n \to \infty}\frac{ln(n)}{n} \\\\ln(L)=5.6\lim_{n \to \infty}\frac{1/n}{1} \\\\ln(L)=5.6*0=0\\\\L=e^{0}=1[/tex]
Therefore, the limit value of given sequece is 1.
Learn more about the limit of function here:
https://brainly.com/question/2166212