Sequences — Question 1

*Question 1 Let $a_n=\dfrac{3n-1}{n+2}$ for $n\ge1$.

1. Find $L=\lim_{n\to\infty}a_n$.

2. Solve $|a_n-L|<0.01$ exactly and determine the smallest integer $N$ such that the inequality holds for every $n\ge N$.

*Question 1 – Solution

The numerator and denominator have the same degree. Divide both by $n$: $$a_n=\frac{3-1/n}{1+2/n}\longrightarrow\frac31=3.$$ Since $1/n\to0$ and $2/n\to0$, the quotient law gives $L=3$.

Rather than approximate numerically, subtract the limit and simplify: $$|a_n-3|=\left|\frac{3n-1}{n+2}-3\right|=\frac7{n+2}.$$

Therefore $$\frac7{n+2}<0.01=\frac1{100}\quad\Longleftrightarrow\quad n>698.$$ Thus every integer $n\ge699$ satisfies the requirement.

The smallest possible integer is $N=699$. At the preceding index, $n=698$, the error is $7/700=0.01$, which equals the tolerance and is not strictly less than it. Hence no smaller $N$ works.