How do you solve for N directly from the error inequality $\frac{9}{n^2-3} < \varepsilon$ ?

Q: How do you solve for N directly from the error inequality $\frac{9}{n^2-3} < \varepsilon$?

To solve directly, isolate $n$. First, note that for the error to be defined and positive, we typically consider $n > \sqrt{3}$. Rearrange the inequality: $n^2 - 3 > \frac{9}{\varepsilon}$. Add 3 to both sides: $n^2 > \frac{9}{\varepsilon} + 3$. Take the square root: $n > \sqrt{\frac{9}{\varepsilon} + 3}$. The integer cutoff $N$ is any integer greater than or equal to this value (strictly, $N$ must be such that $n > N$ implies the condition, so $N = \lfloor \sqrt{\frac{9}{\varepsilon} + 3} \rfloor$ or similar depending on strictness conventions, but the video emphasizes choosing an integer guaranteeing the condition).

To solve directly, isolate

n

. First, note that for the error to be defined and positive, we typically consider

n > \sqrt{3}

. Rearrange the inequality:

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

. Add 3 to both sides:

n^2 > \frac{9}{\varepsilon} + 3

. Take the square root:

n > \sqrt{\frac{9}{\varepsilon} + 3}

. The integer cutoff

N

is any integer greater than or equal to this value (strictly,

N

must be such that

n > N

implies the condition, so

N = \lfloor \sqrt{\frac{9}{\varepsilon} + 3} \rfloor

or similar depending on strictness conventions, but the video emphasizes choosing an integer guaranteeing the condition).

Conditions

$\varepsilon > 0$ .
$n > \sqrt{3}$ to ensure the denominator is positive.
Solving $\frac{9}{n^2-3} < \varepsilon$ .

Reasoning, step by step

Start with $\frac{9}{n^2-3} < \varepsilon$ .
Assume $n^2 - 3 > 0$ (i.e., $n > \sqrt{3}$ ).
Multiply both sides by $n^2 - 3$ and divide by $\varepsilon$ : $\frac{9}{\varepsilon} < n^2 - 3$ .
Add 3 to both sides: $\frac{9}{\varepsilon} + 3 < n^2$ .
Take the positive square root: $\sqrt{\frac{9}{\varepsilon} + 3} < n$ .
Choose an integer $N$ such that $N \ge \sqrt{\frac{9}{\varepsilon} + 3}$ (or strictly greater depending on definition of $n>N$ ).
Verify that this $N$ satisfies the original inequality for all $n > N$ .

Example

The script states: 'For n>√3 the error is 9/(n²−3). Solving the error inequality directly gives n>√( $9/ε+3$ ). Choose an integer cutoff guaranteeing this condition.'

Common misconceptions

Forgetting the domain restriction $n > \sqrt{3}$ .
Incorrectly handling the inequality direction when multiplying by a variable expression (must ensure positivity).
Assuming the direct solution is always computationally simpler than scaling methods (it involves more complex arithmetic).

Watch the explanation

BilibiliWhy the cutoff N in a limit proof is not unique

0:17 – 0:32Watch this moment ↗

BilibiliVisualizing the epsilon-N definition

0:13 – 0:29Watch this moment ↗

Connected concepts

Limits →

How do you solve for N directly from the error inequality $\frac{9}{n^2-3} < \varepsilon$ ?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

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Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

Explore next

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How do you solve for N directly from the error inequality $\frac{9}{n^2-3} < \varepsilon$ ?