A calculus problem by Anupam Khandelwal

From $x_n = \dfrac {4n-5}{9n+2} = \dfrac {\frac 49\left(9n +2 - \frac {53}4 \right)}{9n+2} = \dfrac 49 - \dfrac {53}{9(9n+2)}$ . $\implies \displaystyle \lim_{n \to \infty} x_n = \dfrac 49 = a$ . Therefore, $x_n < a$ for all $n$ . For $x_n < a-0.1$ we have:

$\begin{aligned} \frac {53}{9(9n+2)} & > 0.1 \\ 530 & > 81n + 18 \\ n & < \frac {512}{81} \le 6 \end{aligned}$ .

Therefore, $z=6$ and $2z=\boxed{12}$ .

Given general term of the sequence, $x_n=\frac{4n-5}{9n+2}$ . Then, according to the problem, we have,

$\displaystyle a=\lim_{n\to \infty} x_n = \lim_{n\to \infty} \bigg( \frac{4n-5}{9n+2} \bigg)$

Taking $m=\frac{1}{n}$ , we get that $n\to \infty \implies \frac{1}{n}\to 0 \implies m\to 0$

So, the limit now becomes,

$\displaystyle a=\lim_{m\to 0} \bigg( \frac{4-5m}{9+2m} \bigg)$ [since $m\neq 0$ ]

$\implies a=\frac{4-0}{9+0}=\frac{4}{9}$

So, the given open interval in the problem is $(a-0.1,a+0.1)$ , i.e., $(\frac{31}{90},\frac{49}{90})$

Since, the sequence converges to $a$ , so there can never be any term of the sequence outside the interval $(\frac{31}{90},\frac{49}{90})$ on the right side. For the left side though, let us assume that $\frac{31}{90}$ is the $t^{\text{th}}$ term of the sequence, i.e., $x_t=\frac{31}{90}$ .

From that and using the given expression for the general term of the sequence, we can find that $t\approx 6.321$ , which implies that $\frac{31}{90}$ is not in the sequence and that there are $6$ terms in the left side outside the given interval, so we have $z=6$ .

Then, Reqd. value $=2z=\boxed{12}$