Proof of an Limit

Question

Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$.

Workings:

If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$

If $n$ is large enough $4n^2 + 2$ behaves like $4n^2$

More formally we can find $a,b$ such that $\frac{3n^2+5n}{4n^2 +2} \leq \frac{a}{b} \frac{3n^2}{4n^2}$

For $n\geq 2$ we have $3n^2 + 5n /leq 3n^2.

For $n \geq 0$ we have $4n^2 + 2 \geq \frac{1}{2}4n^2$

So for $ n \geq \max\{0,2\} = 2$ we have:

$\frac{3n^2+5n}{4n^2 +2} \leq \frac{2 \dot 3n^2}{\frac{1}{2}4n^2} = \frac{3}{4}$

To make $\frac{3}{4}$ less than $\epsilon$:

$\frac{3}{4} < \epsilon$, $\frac{3}{\epsilon} < 4$

Take $N = \frac{3}{\epsilon}$

Proof:

Suppose that $\epsilon > 0$

Let $N = \max\{2,\frac{3}{\epsilon}\}$

For any $n \geq N$, we have that $n > \frac{3}{\epsilon}$ and $n>2$, therefore

$3n^2 + 5n^2 \leq 6n^2$ and $4n^2 + 2 \geq 2n^2$

Then for any $n \geq N$ we have

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

$ = \frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}$

$ = \frac{10n-3}{8n^2+4}$

Now I'm not sure on what to do. Any help will be appreciated.

Answer 1

Note

$$\left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right| = \left|\frac{20n - 6}{4(4n^2 + 2)}\right| = \frac{|10n - 3|}{4(2n^2 + 1)} < \frac{10n + 3}{8n^2} < \frac{20n}{8n^2} = \frac{5}{2n}.\tag{1}$$

Hence, given $\epsilon > 0$, setting $N > 2\epsilon/5$ will make the left hand side of $(1)$ less than $\epsilon$ whenever $n \ge N$. Therefore

$$\lim_{n\to \infty} \frac{3n^2 + 5n}{4n^2 + 2} = \frac{3}{4}.$$

Answer 2

It simple

$$ \lim_{x \to \infty} \dfrac{3 + \dfrac{5}{n}}{4 + \dfrac{2}{n^2}}$$

$$ = \dfrac{3}{4}$$

Answer 3

Perhaps simpler:

With the Squeeze Theorem:

$$\frac34\xleftarrow[x\to\infty]{}\frac{3n^2}{4n^2}\le\frac{3n^2+5n}{4n^2+2}\le\frac{3n^2+5n}{4n^2}=\frac34+\frac54\frac1{n}\xrightarrow[n\to\infty]{}\frac34+0=\frac34$$

With arithmetic of limits:

$$\frac{3n^2+5n}{4n^2+2}=\frac{3n^2+5n}{4n^2+2}\cdot\frac{\frac1{n^2}}{\frac1{n^2}}=\frac{3+\frac5n}{4+\frac2{n^2}}\xrightarrow[n\to\infty]{}\frac{3+0}{4+0}=\frac34$$