$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

By using the triangle inequality we can simplify expressions when we have negative signs in the expression for the sequence.

**Example**

Show that $a_n=\frac{2n^4+5n^3-3n^2+1}{n^4} \to 2$.

Again we simplify $|a_n-a|$ and find expressions that it is less than or equal to:

\begin{align*}

|a_n-2|&= \left| \frac{2n^4+5n^3-3n^2+1}{n^4} – 2 \right|

= \left| \frac{5n^3-3n^2+1}{n^4} \right| \\

&\leq \left| \frac{5n^3}{n^4} \right| + \left|\frac{-3n^2}{n^4} \right| + \left| \frac{1}{n^4} \right|, {\text{ by the triangle inequality}},\\

&= \frac{5n^3+3n^2+1}{n^4} \\

&\leq \frac{5n^3+3n^3+n^3}{n^4} , {\text{ for }} n\geq 1, ({\text{as }} n^2\leq n^3 {\text{ and }} 1\leq n^3),\\

&= \frac{9n^3}{n^4} \\

&=\frac{9}{n} .

\end{align*}

Hence we should take $N$ to be an integer greater than $9/\varepsilon $.

As usual let’s carefully write a coherent answer:

**Polished version: **

Show that $a_n=\frac{2n^4+5n^3-3n^2+1}{n^4} \to 2$.

Answer: Given $\varepsilon >0$, let $N$ be an integer greater than $9/\varepsilon$. Then for all $n>N$ we have $n>9/\varepsilon $, i.e, $\varepsilon >9/n$ and hence

\begin{align*}

|a_n-2|&= \left| \frac{2n^4+5n^3-3n^2+1}{n^4} – 2 \right|

= \left| \frac{5n^3-3n^2+1}{n^4} \right| \\

&\leq \left| \frac{5n^3}{n^4} \right| + \left|\frac{-3n^2}{n^4} \right| + \left| \frac{1}{n^4} \right|, {\text{ by the triangle inequality}},\\

&= \frac{5n^3+3n^2+1}{n^4} \\

&\leq \frac{5n^3+3n^3+n^3}{n^4} , {\text{ as }} n\geq 1,\\

&= \frac{9n^3}{n^4} \\

&=\frac{9}{n} \\

&\lt \varepsilon .

\end{align*}