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Example

Prove $\lim \frac{ 1 }{ n^2 } = 0 $.

Proof [expand]

Fix $\epsilon > 0 $, want to find $N $ s.t. $n > N \implies \abs{ \frac{ 1 }{ n^2 } - 0 } < \epsilon$.

$$\begin{align*} \abs{ \frac{ 1 }{ n^2 } - 0 } &< \epsilon \br \frac{ 1 }{ n^2 } &< \epsilon \br n^2 &> \frac{ 1 }{ \epsilon } \br \end{align*}$$

Since $n > 0$, $n > \frac{ 1 }{ \sqrt[ ]{ \epsilon }} $.

Choose $N = \ceil{ \frac{ 1 }{ \sqrt[ ]{ \epsilon }}}$, then $n > N, \abs{ \frac{ 1 }{ n^2 } - 0 } < \epsilon$.

Example

Prove $\lim \frac{ 4n^3 + 3n }{ n^3 - 6 } = 4 $.

Proof [expand]

Fix $\epsilon > 0 $, want to find $N $ s.t. $n > N \implies \abs{ \frac{ 4n^3 + 3n }{ n^3 - 6 } - 4 } < \epsilon $.

$\abs{ \frac{ 3n + 24 }{ n^3 - 6 }} < \epsilon$

let $n > 2$,

$\frac{ 3n + 24 }{ n^3 - 6 } < \epsilon$

We scale the left hand side to be:

$\frac{ 3n + 24 }{ n^3 - 6 } < \frac{ 27n }{ \frac{ 1 }{ 2 } n^3 }$

to make this true, $\begin{cases} 27n > 3n+24 \br \frac{ 1 }{ 2 }n^3 < n^3 -6 \end{cases}$

we get $\begin{cases} n > 1 \br n > \sqrt[ 3 ]{ 12 } \end{cases}$

i.e. along with previous $n > 2$, we choose $n > 3$.

Now we check $\frac{ 27n }{ \frac{ 1 }{ 2 }n^3 } < \epsilon $

$n > \frac{ \sqrt[ ]{ 54 }}{ \epsilon } $.

When $N = \max{ \set{ 3, \ceil{ \frac{ \sqrt[ ]{ 54 }}{ \epsilon }}}}$, we have > N \implies \abs{ \frac{ 4n^3 + 3n }{ n^3 - 6 } - 4 } < \epsilon $.