Sequences — Question 10

Question 10 – Solution

Step 1: Find the limit by continuity.

Since $1/n\to0$ and cosine is continuous at $0$, $$L=\lim_{n\to\infty}\cos(1/n)=\cos0=1.$$

Step 2: Express and bound the error.

Because $\cos(1/n)\le1$, the absolute error simplifies to $$|a_n-L|=1-\cos(1/n).$$ Apply $0\le1-\cos x\le x^2/2$ with $x=1/n$: $$0\le |a_n-1|\le\frac{1}{2n^2}.$$

Step 3: Solve for a sufficient index.

It is enough to make the upper bound smaller than the desired tolerance: $$\frac1{2n^2}<10^{-4} \quad\Longleftrightarrow\quad n^2>5000 \quad\Longleftrightarrow\quad n>\sqrt{5000}\approx70.71.$$

Step 4: State the guarantee precisely.

Thus $N=71$ guarantees $|a_n-1|<10^{-4}$ for every $n\ge71$, because $1/(2n^2)$ decreases with $n$. This is a sufficient value obtained from an upper bound; it need not be the smallest possible $N$ for the exact cosine error.