How does the scaling technique $n^2-3 \ge n^2/2$ simplify the error bound for $n \ge 3$ ?

Q: How does the scaling technique $n^2-3 \ge n^2/2$ simplify the error bound for $n \ge 3$?

For $n \ge 3$, the inequality $n^2 - 3 \ge n^2/2$ holds. This allows replacing the complex denominator $n^2 - 3$ in the error term $\frac{9}{n^2-3}$ with the simpler $n^2/2$. Since the denominator is made smaller (or equal), the fraction becomes larger (or equal), providing an upper bound: $\frac{9}{n^2-3} \le \frac{9}{n^2/2} = \frac{18}{n^2}$. Solving $\frac{18}{n^2} \sqrt{18/\varepsilon}$, which is easier to compute than the direct algebraic solution.

For

n \ge 3

, the inequality

n^2 - 3 \ge n^2/2

holds. This allows replacing the complex denominator

n^2 - 3

in the error term

\frac{9}{n^2-3}

with the simpler

n^2/2

. Since the denominator is made smaller (or equal), the fraction becomes larger (or equal), providing an upper bound:

\frac{9}{n^2-3} \le \frac{9}{n^2/2} = \frac{18}{n^2}

. Solving

\frac{18}{n^2} < \varepsilon

yields

n > \sqrt{18/\varepsilon}

, which is easier to compute than the direct algebraic solution.

Conditions

$n \ge 3$ .
Error term is $\frac{9}{n^2-3}$ .
Goal is to find $N$ such that error $< \varepsilon$ .

Reasoning, step by step

Verify the inequality $n^2 - 3 \ge n^2/2$ for $n \ge 3$ .
Apply this to the denominator of the error term: since $n^2 - 3 \ge n^2/2$ , then $\frac{1}{n^2 - 3} \le \frac{1}{n^2/2}$ .
Multiply by the numerator 9 to get the bound: $\frac{9}{n^2-3} \le \frac{18}{n^2}$ .
Set the simplified bound less than $\varepsilon$ : $\frac{18}{n^2} < \varepsilon$ .
Solve for $n$ : $n^2 > \frac{18}{\varepsilon} \implies n > \sqrt{\frac{18}{\varepsilon}}$ .
Choose an integer $N$ greater than this value.

Example

The script states: 'For $n\ge 3$ , n²− $3\ge n$ ²/2, so the error is bounded by $18/n$ ². Requiring n>√( $18/ε$ )... gives another valid cutoff.'

Common misconceptions

Assuming the inequality direction reverses when taking reciprocals without checking signs (here terms are positive).
Forgetting that this bound is only valid for $n \ge 3$ , so the final $N$ must also satisfy this range restriction.
Believing this method yields the exact minimal $N$ ; it yields a valid, possibly larger, $N$ .

Watch the explanation

BilibiliWhy the cutoff N in a limit proof is not unique

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Connected concepts

Limits →

How does the scaling technique $n^2-3 \ge n^2/2$ simplify the error bound for $n \ge 3$ ?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

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Related questions

How does the scaling technique n2−3≥n2/2n^2-3 \ge n^2/2n2−3≥n2/2 simplify the error bound for n≥3n \ge 3n≥3?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

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How does the scaling technique $n^2-3 \ge n^2/2$ simplify the error bound for $n \ge 3$ ?