Prove the following:

$$\lim_{n\to \infty}\frac{6n^2+3n}{2n^2-5}=3$$

Here's my solution:

Consider the function $\frac{6n^2+3n}{2n^2-5}$.

Then $$|\frac{6n^2+3n}{2n^2-5}-3|= |\frac{6n^2+3n}{2n^2-5}-\frac{6n^2+15}{2n^2-5}|=|\frac{3n-15}{2n^2-5}|$$

The upper bound for the numerator is $3n-15 \le 3n$ for all $n$ and the lower bound for the denominator is $2n^2-5 \le 4n^2$ if $n \le 2$.

It follows that

$|\frac{3n-15}{2n^2-5}| \le \frac{3n}{4n^2}$. Meaning $|\frac{3n-15}{2n^2-5}| \le \frac{3}{4n}$.

By the Archimedean Property, $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$

Therefore $(\frac{6n^2+3n}{2n^2-5})\rightarrow 3$.


I feel like I'm missing a step after showing that $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$.

Can $\frac{3}{4n} \le \frac{3}{4N} \lt \epsilon$ be simplified more so that it resembles $\frac{1}{n} \le \frac{1}{N} \lt \epsilon$ or is it fine how it is?

Please advise.

  • $\begingroup$ "and the lower bound for the denominator is $2n^2 - 5 \leq 4n^2$ if $n \leq 2$"... I think you mean $n \geq 2$ $\endgroup$ – amarney Oct 27 '17 at 1:26
  • $\begingroup$ Once you have established that $|a_n - L| \leq \frac{3}{4n}$ for $n \geq 2$, you should be able to see that taking $n$ to be greater than $N = \frac{3}{4\epsilon}$ gets you within $\epsilon$. You can think of proving the convergence of a sequence as completing the following challenge: Given any $\epsilon > 0$, can you find an integer $N$ large enough? $\endgroup$ – amarney Oct 27 '17 at 1:31

$$\left|\frac{6n^2+3n}{2n^2-5}-3\right|= \left|\frac{6n^2+3n}{2n^2-5}-\frac{6n^2\color{red}-15}{2n^2-5}\right|=\left|\frac{3n\color{red}+15}{2n^2-5}\right|$$

Hence upper bound for the numerator when $n> 15$, is $|3n+15| \leq 4n.$

A lower bound for the denominator for $n \geq 3$, is $|2n^2-5| \geq n^2$

It follows that $\left|\frac{3n+15}{2n^2-5}\right|\leq \frac4n$ for $n>15$.

Now, by archimedian property, $\forall \epsilon>0 , \exists N>0, \frac{4}{N}<\epsilon$, now let $n>\max(N, 15)$. and we have

$$\left|\frac{3n+15}{2n^2-5}\right|\leq \frac4n \leq \frac4N<\epsilon$$


To get an upper bound, you want to replace the numerator by something big and the denominator by something small. So instead of $2n^2-5\lt4n^2$, you should use $2n^2-5\gt2n^2$ (or even just $n^2$). Aside from that, everything is basically OK.


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