Proving Limit Using Epsilon Criteria: n^2+1/n+3 → +∞ as n → ∞

[squareroot]

Homework Statement

Prove the limit by using the ε criteria
$\stackrel{lim}{n→∞}$$\frac{n^2+1}{n+3}$ = +∞

Homework Equations

The Attempt at a Solution

Hi guys.So I'm having trouble proving this limits(all limits that i need to prove that they equal +/- ∞) , all that I've learned in calculus so far at school is these e proprieties

limit when x→∞ from x_n=L ⇔ $\forall$ε>0 $\exists$n_ε so that for $\forall$ n≥n_ε we have |x_n-L|<ε

the second one is when limit whenx→∞ from x_n= +∞ it's the same except the last part where you don't have |x_n-L|<ε but we have x_n≥ε

fot the third one limit when x→∞ from x_n= -∞ when x_n≤-ε

TY!

 

[Mark44]

Homework Statement

Prove the limit by using the ε criteria
$\stackrel{lim}{n→∞}$$\frac{n^2+1}{n+3}$ = +∞

Homework Equations

The Attempt at a Solution

Hi guys.So I'm having trouble proving this limits(all limits that i need to prove that they equal +/- ∞) , all that I've learned in calculus so far at school is these e proprieties

limit when x→∞ from x_n=L ⇔ $\forall$ε>0 $\exists$n_ε so that for $\forall$ n≥n_ε we have |x_n-L|<ε

the second one is when limit whenx→∞ from x_n= +∞ it's the same except the last part where you don't have |x_n-L|<ε but we have x_n≥ε

fot the third one limit when x→∞ from x_n= -∞ when x_n≤-ε

TY!

You're not using the correct form for this type of limit. This is the one you want.
$$ \forall M, \exists N \text{ such that } n \ge N \Rightarrow f(n) > M$$

The idea here is that M and N are large numbers. You don't use $\epsilon$ in this type of limit proof.

 

[squareroot]

well what you said is the same with what i said "the second one is when limit whenx→∞ from xn= +∞ it's the same except the last part where you don't have |xn-L|<ε but we have xn≥ε
"

 

[Mark44]

well what you said is the same with what i said "the second one is when limit whenx→∞ from xn= +∞ it's the same except the last part where you don't have |xn-L|<ε but we have xn≥ε
"

Which means that it isn't the same. Also, $\epsilon$ is generally considered to be a small number (close to 0).

 

[squareroot]

Well yes but i tried solving it like that and i get lost in all that math , i have a polynominal equation in n...can't you show me?

 

[Mark44]

The way PF works is that we help you, but we don't do your work for you. The polynomial is a quadratic, and you should already know how to solve quadratic equations and inequalities. Show us where you got stuck.

 

[squareroot]

Ok here we go , so starting backwards from the theorem we have (n²+1)/(n+3)≥ε after solving this and by arranging it you have

n²-nε+(1-3ε)≥0

from here

Δ=ε²-4(1-3ε)=ε²+12ε-4




so it from here where i m not certain what to do i tried to put the condition that Δ≥0 but by solving that i get really ugly results.I don't know what to do from here!

 

[STEMucator]

You want :

$\forall M > 0, \exists N > 0 \space | \space n>N \Rightarrow a_n > M$

So start by massaging what you want into a suitable form :

$\frac{n^2 + 1}{n + 3} > M$

 

[Mark44]

You should not use ε for this type of limit, as it is almost universally used to represent numbers that are close to 0.

From the inequality $$\frac{n^2+1}{n + 3} > M$$
you get
$n^2 - Mn + 1 - 3M > 0$
The values of n that make the expression above equal to 0 are
$$n = \frac{M \pm \sqrt{M^2 + 12M - 4}}{2}$$

It is not necessary to get an exact value for n. All that is needed is to find a value that is large enough. $\sqrt{M^2 + 12M - 4}$ ≈ M, but the square root expression is always a little larger. I think we can say that the radical is smaller than 2M as long as M is not too small. For example, if M = 10 (which isn't very large), $\sqrt{M^2 + 12M - 4}=\sqrt{216}≈11 < 20 = 2M$. For any larger value of M the radical will be smaller than 2M.

Can you carry on from here?

 

[squareroot]

Not really , the thing is i don't really have a solid understanding about what am I after to prove here.So we've proved that that radical will be smaller than 2M when M is a large number, but hor does that help me with my initial inequation ?I'm sorry i know I'm pretty much asking to solveit step by step to me , but it 's my first kind of this exercise from my first calculus homework ever , and i really want to have a solid foundation on calculus.

 

[Dick]

Not really , the thing is i don't really have a solid understanding about what am I after to prove here.So we've proved that that radical will be smaller than 2M when M is a large number, but hor does that help me with my initial inequation ?I'm sorry i know I'm pretty much asking to solveit step by step to me , but it 's my first kind of this exercise from my first calculus homework ever , and i really want to have a solid foundation on calculus.

You can make life a lot easier by simplifying the problem first. $\frac{n^2 + 1}{n + 3} > \frac{n^2}{n + 3} > \frac{n^2}{n + n}$ (for n>3). Simplifying the last expression and showing its limit is infinity shows the first one has that limit as well.

 

[squareroot]

Oh!I get it now! And it's obvious that n²/2n's limit is +∞ so from that i can conclude that the limit of my equation is also +∞.This is very very useful , and i can use it to solve all my exercises of this type.Thanks a lot! But is it a valid method? Seems more of an observation than a demonstration to me

 

[squareroot]

For example here is how i could solve another exercise which looks similar

Prove that lim_n→∞[itex]\frac{n³+2n}{n²+3n-1}[/itex]=+∞

So [itex]\frac{n³+2n}{n²+3n-1}[/itex] < [itex]\frac{n³+2n}{n²+3n}[/itex]
But [itex]\frac{n³+2n}{n²+3n}[/itex]=[itex]\frac{n(n²)+2}{n(n+3)}[/itex]=[itex]\frac{n²+2}{n+3}[/itex], now here 2 and 3 are pretty irrelevant when n→∞ so we could say [itex]\frac{n²}{n}[/itex]=n
lim_n→∞ n=+∞, that is obvious so from this i proved that lim_n→∞[itex]\frac{n³+2n}{n²+3n-1}[/itex]=+∞

What do you say?Is it ok?

P.S idk why the eq editor didn't work!

 

[Dick]

Oh!I get it now! And it's obvious that n²/2n's limit is +∞ so from that i can conclude that the limit of my equation is also +∞.This is very very useful , and i can use it to solve all my exercises of this type.Thanks a lot! But is it a valid method? Seems more of an observation than a demonstration to me

You can make it into a real proof. If n^2/(2n)>M then (n^2+1)/(n+3)>M. You can use the same bound for the simpler expession for the more complicated.

 

[micromass]

For example here is how i could solve another exercise which looks similar

Prove that lim_n→∞[itex]\frac{n³+2n}{n²+3n-1}[/itex]=+∞

So [itex]\frac{n³+2n}{n²+3n-1}[/itex] < [itex]\frac{n³+2n}{n²+3n}[/itex]
But [itex]\frac{n³+2n}{n²+3n}[/itex]=[itex]\frac{n(n²)+2}{n(n+3)}[/itex]=[itex]\frac{n²+2}{n+3}[/itex], now here 2 and 3 are pretty irrelevant when n→∞ so we could say [itex]\frac{n²}{n}[/itex]=n
lim_n→∞ n=+∞, that is obvious so from this i proved that lim_n→∞[itex]\frac{n³+2n}{n²+3n-1}[/itex]=+∞

What do you say?Is it ok?

P.S idk why the eq editor didn't work!

Do not use [NOPARSE][/NOPARSE] and [itеx][/itеx] together, if you want to write supscripts, then please write

[NOPARSE][itex]n^3+2n[/itex][/NOPARSE]

Also, limits can be written nicely as

[NOPARSE]
[itex]\lim_{n\rightarrow +\infty} n^2[/itex]
[/NOPARSE]

Here is a LaTeX guide:

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