I dont know how i can resolve this problem of the ε definition of a limit of succession

Question

I have this two problem to resolve with the ε-δ definition of a limit of succession $$\lim_{n\to\infty} \frac{2n+3}{3n-50}=\frac{2}{3}$$ $$lim_{n\to\infty} \frac{3n^2-12n+1}{n+25}=+\infty$$ With the first one $$\lvert \frac{2n+3}{3n-50} \rvert \lt \varepsilon$$ $$\lvert \frac{-4n+103}{3(3n-50)} \rvert \lt \varepsilon$$ $$\frac{1}{3}\lvert \frac{-4n+103}{3n-50} \rvert \lt \varepsilon$$ $$ \frac{4n+103}{3n+50} \lt 3\varepsilon$$ but from this point i dont know what to do

And with the second problem i dont even know where to start because its $\infty$ so i can't take that from the equation. Can someone explain what can i do to resolve this? Or give a small explanation of how to resolve this type of problems?

Answer 1

For the first one, you have to prove that, given $\ \varepsilon>0,\ \exists\ N\in\mathbb{N}\ $ such that $$\ \left\vert\frac{2n+3}{3n-50} - \frac{2}{3}\right\vert\ < \varepsilon\quad \forall\ n\geq N.$$

So consider the fact that

$$\frac{2n+3}{3n-50} - \frac{2}{3} = \frac{6n+9}{9n-150} - \left(\frac{6n-100}{9n-150} \right) = \frac{109}{9n-150}\ < \frac{109}{n}\quad \forall n\geq 19.$$

Therefore, for all $\ n\geq 19,\ $

$$ 0 < \left\vert\frac{2n+3}{3n-50} - \frac{2}{3}\right\vert = \frac{2n+3}{3n-50} - \frac{2}{3} = \frac{109}{9n-150} < \frac{109}{n},$$

and you should be able to use the Archimedean property of real numbers to complete the $\ \varepsilon-n\ $ proof.

$$$$

For the second one, you must show that given $\ x\in\mathbb{R},\ \exists\ N\in\mathbb{N}\ $ such that $$\ \frac{3n^2-12n+1}{n+25} > x\quad \forall n\geq N.\ $$

To do this, just use polynomial long division and then show that you can find an $\ N\ $ that does this for each $\ x.$

Answer 2

Your mistake is in the definition of limit of a sequence. Let $s_n$ be a sequence of real numbers such as $\frac{2n+3}{3n-50}$, then the sequence converges if there exists $l\in\mathbb{R}$ such that $\lim_{n\rightarrow\infty}s_n=l$, which exactly means this:

$$ \exists l\in\mathbb{R}\forall \epsilon \in\mathbb{R^+}\exists N \in\mathbb{N}\forall n\in\mathbb{N}: n\geq N \Rightarrow |s_n-l|<\epsilon $$

In the first part of your exercise you forgot to put $\frac{2}{3}$ inside the modulus. Try to do this and simplify.

The definition for infinite limits is in part similar, and transmits the idea that the modulus of sequence $|s_n|$ can get arbitrarly bigger from certain order. For example $\frac{3n^2-12n+1}{n+25}=3n-87+\frac{2176}{n+25}$ is greater than a million from $n>333363$. An algebric way of solving the second problem is by confirming that $\frac{2176}{n+25}$ converges to zero and $3n-87$ diverges.