Prove that $\lim_{n \to \infty} \frac{2n^2}{5n^2+1}=\frac25$

Question

Prove that $\lim_{n \to \infty} \dfrac{2n^2}{5n^2+1}=\dfrac{2}{5}$

$$\forall \epsilon>0 \ \ :\exists N(\epsilon)\in \mathbb{N} \ \ \text{such that} \ \ \forall n >N(\epsilon) \ \ \text{we have} |\dfrac{-2}{25n^2+5}|<\epsilon$$ $$ |\dfrac{-2}{25n^2+5}|<\epsilon \\ |\dfrac{2}{25n^2+5}|<\epsilon \\ \dfrac{5(n^2+1)}{2}<\epsilon \\ {5(n^2+1)}<2\epsilon \\ {n}>\sqrt{\dfrac{2\epsilon -5}{25}}$$

What do I do?

Answer 1

Given the limit $\lim_{n \to \infty} \dfrac{2n^2}{5n^2+1}$ you have to group by $n^2$. It becomes: $$\lim_{n \to \infty} \dfrac{2n^2}{5n^2+1}=\lim_{n \to \infty} \dfrac{n^2\cdot 2}{n^2\cdot\left(5+\frac 1{n^2}\right)}=\lim_{n \to \infty} \dfrac{2}{\left(5+\frac 1{n^2}\right)}=\dfrac 25$$

Because we know that $\lim_{n\to \infty}\left(5+\frac 1{n^2}\right)=5$

Answer 2

$$\forall \epsilon>0 \ \ :\exists n_0\in \mathbb{N} \ \ \text{such that} \ \ \forall n >n_0 \quad \left|\dfrac{2n^2}{5n^2+1}-\dfrac{2}{5}\right|<\epsilon$$

$$\left|\dfrac{-2}{25n^2+5}\right|<\epsilon\iff \dfrac{2}{25n^2+5}<\epsilon\iff 25n^2+5>\frac{2}{\epsilon}\iff25n^2> \frac{2}{\epsilon}-5\iff n^2>\frac{2}{25\epsilon}-\frac15\implies n> \sqrt{\frac{2}{25\epsilon}-\frac15}$$

Answer 3

Note that $\frac{2n^2}{5n^2+1} = \frac{2}{5} \frac{n^2}{n^2+\frac{1}{5}}$. Subtract $\frac{2}{5}$, and you get: $\frac{2}{5} \left|\frac{n^2}{n^2+\frac{1}{5}}-1\right| = \frac{2}{5}\left|\frac{\frac{1}{5}}{n^2+\frac{1}{5}}\right|$. Now we can always take $N>0$ such that for all $n>N$, this is smaller than $\epsilon$. Can you see why?

Answer 4

Let $\varepsilon >0$

$|\dfrac{-2}{25n^2+5}|<\varepsilon\iff \dfrac{2}{25n^2+5}<\varepsilon$

$\dfrac{2}{25n^2+5}<\dfrac{2}{25n^2}<\varepsilon$ so $n^2>\dfrac{2}{25\varepsilon}\iff n>\dfrac{\sqrt{2}}{5\sqrt{\varepsilon}}$

Then $N(\varepsilon)=\bigg\lfloor\dfrac{\sqrt{2}}{5\sqrt{\varepsilon}}\bigg\rfloor+1$

Answer 5

$|{\frac{2}{5(5n^2+1)}}|<\frac{2}{25n^2}<\varepsilon$. So $n>\frac{\sqrt2}{5\sqrt\varepsilon}$ and $N=[{\frac{\sqrt2}{5\sqrt\varepsilon}}]+1$

Answer 6

If you want an $\epsilon,\delta$ proof of the limit then given an $\epsilon>0$ you must find an $N>0$ such that if $n>N$ then

$$ \left\vert \frac{2n^2}{5n^2+1}-\frac{2}{5}\right\vert<\epsilon $$

That is,

$$ \frac{2}{25n^2+5}<\epsilon $$

You may replace the expression on the left by a larger expression to simplify the exercise, such as for example

$$\frac{2}{25n^2+5}<\frac{2}{25n^2}<\frac{1}{n^2}$$

Then if you can find for which values of $n$ it is true that

$$ \frac{1}{n^2}<\epsilon $$ then it will also be true that

$$ \frac{2}{25n^2+5}<\epsilon $$

So clearly we want $n>N>\dfrac{1}{\sqrt{\epsilon}}$.

But this analysis is just the prequel to the actual proof.

$$\text{Prove that } \lim_{n \to \infty} \dfrac{2n^2}{5n^2+1}=\dfrac{2}{5} $$

Let $\epsilon>0$ and let $N>\dfrac{1}{\sqrt{\epsilon}}$. Then $\frac{1}{N^2}<\epsilon$. Let $n>N$. Then

\begin{eqnarray} \left\vert \frac{2n^2}{5n^2+1}-\frac{2}{5}\right\vert&=&\frac{2}{25n^2+5}\\ &<&\frac{1}{n^2}\\ &<&\frac{1}{N^2}\\ &<&\epsilon \end{eqnarray}