Another two limits at infinity

[theakdad]

Good! Let's keep the limit notation too:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{4}{6}}+n^{\frac{1}{3}}+2} \)

In the first term in ther denominator, I would reduce the rational exponent:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{4}{3}}+n^{\frac{1}{2}}+1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+2} \)

Now, divide each term in both the numerator and denominator by \(\displaystyle n^{\frac{2}{3}}\). This is the same as multiplying the entire expression of which we are taking the limit by one, so we are only changing its form, not its value.

Let's look at the first term in the numerator so divided:

\(\displaystyle \frac{n^{\frac{4}{3}}}{n^{\frac{2}{3}}}\)

To what does this simplify?

It becomes n² ?

 

[MarkFL]

It becomes n² ?

If it were:

\(\displaystyle n^{\frac{4}{3}}\cdot n^{\frac{2}{3}}\)

then we would add exponents, and we would in fact get $n^2$. But we are dividing, so we want to subtract the exponent in the denominator from the exponent in the numerator. Recall the rule for exponents:

\(\displaystyle \frac{a^b}{a^c}=a^{b-c}\)

So what do you find? Can you now do this to all of the terms? For the constant terms, think of there being a factor of $n^0$. :D

 

[theakdad]

If it were:

\(\displaystyle n^{\frac{4}{3}}\cdot n^{\frac{2}{3}}\)

then we would add exponents, and we would in fact get $n^2$. But we are dividing, so we want to subtract the exponent in the denominator from the exponent in the numerator. Recall the rule for exponents:

\(\displaystyle \frac{a^b}{a^c}=a^{b-c}\)

So what do you find? Can you now do this to all of the terms? For the constant terms, think of there being a factor of $n^0$. :D

Then i think it is \(\displaystyle n^\frac{2}{3}+n^\frac{1}{6}+\frac{1}{2}\)

 

[Petrus]

Then i think it is \(\displaystyle n^\frac{2}{3}+n^\frac{1}{6}+\frac{1}{2}\)

What is \(\displaystyle \frac{1n^0}{n^2}=1n^{0-2}\)
note that \(\displaystyle a^{-b}=\frac{1}{a^b}\)

Regards,
\(\displaystyle |\pi\rangle\)

 

[MarkFL]

Then i think it is \(\displaystyle n^\frac{2}{3}+n^\frac{1}{6}+\frac{1}{2}\)

No, that's incorrect. Is that supposed to represent the numerator of the expression? Can you show me how you arrived at that?

 

[theakdad]

No, that's incorrect. Is that supposed to represent the numerator of the expression? Can you show me how you arrived at that?

First two terms are the results as i have divided them,and third them i was wrong...
So the third therm should be 1^-1 i think so...

 

[Petrus]

First two terms are the results as i have divided them,and third them i was wrong...
So the third therm should be 1^-1 i think so...

This is numerator right?
if so that is not correct! Look what i posted early

Regards,
\(\displaystyle |\pi\rangle\)

 

[theakdad]

This is numerator right?
if so that is not correct! Look what i posted early

Regards,
\(\displaystyle |\pi\rangle\)

Its not the numerator,that what i wrote is what is left when i have divided numerator with denumerator.

 

[MarkFL]

First two terms are the results as i have divided them,and third them i was wrong...
So the third therm should be 1^-1 i think so...

Show me your computations...the third term won't be a constant, it will be a function of $n$.

 

[theakdad]

Show me your computations...the third term won't be a constant, it will be a function of $n$.

Why function of n when there is no n,its only 1?

 

[MarkFL]

Why function of n when there is no n,its only 1?

But when you divide by \(\displaystyle n^{\frac{2}{3}}\), then there will be an $n$ there. :D

 

[theakdad]

But when you divide by \(\displaystyle n^{\frac{2}{3}}\), then there will be an $n$ there. :D

I have lost myself in this limit...

 

[MarkFL]

I have lost myself in this limit...

Okay, this is what I would do, beginning with the point where we converted from radical notation to rational exponents, and reduced all exponents:

Spoiler

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{4}{3}}+n^{\frac{1} {2}}+1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+2}\)

Dividing each term by \(\displaystyle n^{\frac{2}{3}}\) we obtain:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{2}{3}}+n^{-\frac{1} {6}}+n^{-\frac{2}{3}}}{1+n^{-\frac{1}{3}}+2n^{-\frac{2}{3}}}\)

All terms with negative exponents will go to zero an $n\to\infty$ and all terms with positive exponents will go to $\infty$, hence:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{2}{3}}+n^{-\frac{1} {6}}+n^{-\frac{2}{3}}}{1+n^{-\frac{1}{3}}+2n^{-\frac{2}{3}}}=\frac{\infty+0+0}{1+0+0}=\infty\)

 

[Petrus]

What is \(\displaystyle \frac{1n^0}{n^{2/3}}=1n^{0-2/3}\)
note that \(\displaystyle a^{-b}=\frac{1}{a^b}\)

Regards,
\(\displaystyle |\pi\rangle\)

Ok so I see I made a misstake you are going to divide it by \(\displaystyle n^{2/3}\) not \(\displaystyle n^2\) but I Edit it in this post! What you get from this?

 

[theakdad]

Okay, this is what I would do, beginning with the point where we converted from radical notation to rational exponents, and reduced all exponents:

Spoiler

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{4}{3}}+n^{\frac{1} {2}}+1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+2}\)

Dividing each term by \(\displaystyle n^{\frac{2}{3}}\) we obtain:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{2}{3}}+n^{-\frac{1} {6}}+n^{-\frac{2}{3}}}{1+n^{-\frac{1}{3}}+2n^{-\frac{2}{3}}}\)

All terms with negative exponents will go to zero an $n\to\infty$ and all terms with positive exponents will go to $\infty$, hence:

\(\displaystyle \lim_{n\to\infty}\frac{n^{\frac{2}{3}}+n^{-\frac{1} {6}}+n^{-\frac{2}{3}}}{1+n^{-\frac{1}{3}}+2n^{-\frac{2}{3}}}=\frac{\infty+0+0}{1+0+0}=\infty\)

I see where my mistake was... i have divided the numerator with denumerator...instead to divide each term with \(\displaystyle n^\frac{2}{3}\)

As my exam is closer,i make stupid mistakes like that...

 

[Petrus]

As my exam is closer,i make stupid mistakes like that...

None misstake is stupid! Evryone make misstake and it make us think about it so it Will not happened again!
I Hope you understand WHY we divide by the highest power in bottom!
You may want to try check some algebra Rules and try get used with them! Like \(\displaystyle a^2a^1=a^{2+1}\)

Regards,
\(\displaystyle |\pi\rangle\)

 

[theakdad]

None misstake is stupid! Evryone make misstake and it make us think about it so it Will not happened again!
I Hope you understand WHY we divide by the highest power in bottom!
You may want to try check some algebra Rules and try get yard with them! Like \(\displaystyle a^2a^1=a^{2+1}\)

Regards,
\(\displaystyle |\pi\rangle\)

I know Petrus,thank you for reply...
I know algebra rules,but I am so ocupied with work and study,my head is full so sometimes i get lost...

 

[theakdad]

I Hope you understand WHY we divide by the highest power in bottom! Regards,
\(\displaystyle |\pi\rangle\)

Im not sure about that...but i think because we want to simplify the fraction,and then see where each term is showing to...

 

[Petrus]

Im not sure about that...but i think because we want to simplify the fraction,and then see where each term is showing to...

Well the point is that you Will get on bottom is that one Will cancel out each other and rest Will be zero as you can see what happened to this! Remember that divide by highes power in bottom!
Don't try OVER react for exam! That is not smart as you Will just overstrained! Trust me you Will succed your exam! If you got any question or anything you wounder MHB Will help you! Never hessitate to ask any question!

Regards,
\(\displaystyle |\pi\rangle\)

 

[theakdad]

Well the point is that you Will get on bottom is that one Will cancel out each other and rest Will be zero as you can see what happened to this! Remember that divide by highes power in bottom!
Don't try OVER react for exam! That is not smart as you Will just overstrained! Trust me you Will succed your exam! If you got any question or anything you wounder MHB Will help you! Never hessitate to ask any question!

Regards,
\(\displaystyle |\pi\rangle\)

Thank you,im really glad that I am a member of this forum,and thankfull for your help,and help from others!

So the main point of limits is to get as much zeroes you can?

 

[Petrus]

Thank you,im really glad that I am a member of this forum,and thankfull for your help,and help from others!

So the main point of limits is to get as much zeroes you can?

Yeah pretty much! (Remember we are talking about \(\displaystyle x \to \infty\)

Regards,
\(\displaystyle |\pi\rangle\)

 

[theakdad]

Yeah pretty much! (Remember we are talking about \(\displaystyle x \to \infty\)

Regards,
\(\displaystyle |\pi\rangle\)

I have been on the search on the internet where i could find similar problems to mine...but without luck...have you any idea where or how should i "fight" with this limit?

 

[Petrus]

I have been on the search on the internet where i could find similar problems to mine...but without luck...have you any idea where or how should i "fight" with this limit?

read this first it explain well
Pauls Online Notes : Calculus I - Limits At Infinity, Part I

Regards,
\(\displaystyle |\pi\rangle\)

 

[theakdad]

read this first it explain well
Pauls Online Notes : Calculus I - Limits At Infinity, Part I

Regards,
\(\displaystyle |\pi\rangle\)

Will take a look,thank you!