Question

a)Determine whether the sequence is divergent or convergent. If it is convergent, evaluate its limit. If it diverges to infinity, state your answer as INF. If it diverges to negative infinity, state your answer as MINF. If it diverges without being infinity or negative infinity, state your answer as DIV.
lim n goes to infinity -6n^6+sin^2(7n)/(n^7+11)
b) Find the length of the curve defined by y=4x^(3/2)+7 at when x=1 to x=6

Answer 1

Answer:

a) The sequence converges to 0

b) The lenght of the curve is $\frac{1}{54}(217^{3/2}-37^{3/2})$

Step-by-step explanation:

Consider the sequence $a_n = \frac{-6n^6 + \sin^2(7n)}{n^7+11}$

a) We will prove it using the sandwich lemma. Note that for all n $-1\leq \sin^2(7n)\leq 1$, then

$\frac{-6n^6 -1}{n^7+11}\leq\frac{-6n^6 + \sin^2(7n)}{n^7+11}\leq \frac{-6n^6 + 1}{n^7+11}$

Note that the expressions on the left and the right hand side have a greater degree on the denominator than the one on the numerator. Then, by takint the limit n goes to infinty on both sides, we have that

$0 \leq\frac{-6n^6 + \sin^2(7n)}{n^7+11} \leq 0$

So, the sequence converges to 0.

b) The function $f(x) = 4x^{3/2}+7$ the formula of curve lenght is given by

$s = \int_a^b \sqrt[]{1+(f'(x))^2}dx$

in this case, a=1, b=6

Note that $f'(x) =6x^{{1/2}$. Then

$s=\int_1^6 \sqrt[]{1+36x}dx$. Take u  = 1+36x. Then du= 36dx (i.e du/36 = dx). If x = 1, then u = 37 and if x = 6 then u = 217. So,

$s=\frac{1}{36}\int_{37}^{217}\sqrt[]{u} du = \frac{2}{36\cdot 3} \left.u^{3/2}\right|_{37}^{217}=\frac{1}{54}(217^{3/2}-37^{3/2})$