Sequences — Question 1

Question 1 – Solution

Step 1: Find the candidate limit.

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 $L=3$.

Step 2: Express the error exactly.

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

Step 3: Solve the strict accuracy inequality.

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

Step 4: Verify minimality.

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.