Question

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).

Answer 1

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)

$L = \lim_{n \to \infty} b_n \\\\\\L= \lim_{n \to \infty} n^{\frac{5.6}{n} }$

Log on both sides

$In (L) = \lim_{n \to \infty} In (n)^{\frac{5.6}{n} }\\\\= \lim_{n \to \infty} \frac{5.6}{n} In(n)$

$=5.6 \lim_{n \to \infty} \frac{d}{dn} In(n)/\frac{d}{dn} (n)\\\\=5.6 \lim_{n \to \infty} \frac{1}{n} /1 \\\\=5.6 \lim_{n \to \infty} \frac{1}{n} \\\\=5.6 \times 0\\\\In(L) =0\\\\L=e^0\\\\L=1$

$\therefore \lim_{n \to \infty} (n)^{\frac{5.6}{n} =1$