Question

1)

(a) Find the limit of the sequence whose kth term is sirk) Confirm your answer graphically, and show your graph for full credit. (6) Let be K + 4k 3k? + 7 k + 8 If the sequence {b} converges, find its limit.

2)

3)

4)

5)

Answer 1

Solution:a)To find the limit of the sequence whose $k^{th}$ term is , we need to solve the following:

sin(k)

$\lim_{k\rightarrow \infty}\frac{\sin (k)}{k}$

The value of
sin(k)
alternates between -1 and 1 irrespective of the value of  . Therefore, the numerator term i.e.  
sin(k)
of the given series will always have a finite value, whereas the value of denominator term keeps on increasing and finally tends to infinity. Therefore, when the value of is substituted in the given limit, we will get as result. Therefore, we get:

$\boxed{\lim_{k\rightarrow \infty}\frac{\sin (k)}{k}=0}$

In order to verify this graphically, graph the function   using any graphical tool. We will get the following graph:

sin(k)

sin(k) k 4+ 3+ 3 21 1 + k -41 -31 -21 -T 0 TT 21 3л 41 -1 2 -3 -4

As can be seen from the graph above, the value of the given sequence keeps on decreasing progressively as the value of increases and will ultimately die down to zero.

b) Given to us that the sequence $b_k$ converges:

$b_k=\frac{k^2+4k}{3k^2+7k+8}$

In order to find the limit of the given convergent series, we will divide the numerator and the denominator terms by $k^2$ to get:

$\lim_{k\rightarrow \infty}b_k=\lim_{k\rightarrow \infty} \ \frac{1+\frac{4}{k}}{3+\frac{7}{k}+\frac{8}{k^2}}$

Substituting the value of limit, we get:

$=\frac{1+\frac{4}{\infty}}{3+\frac{7}{\infty}+\frac{8}{\infty}}$

$=\frac{1+0}{3+0+0}$

$\Rightarrow \boxed{\lim_{k\rightarrow \infty}=\frac{1}{3}}$