Solve Limits of Sequence: Detailed Instructions

[theakdad]

Im new to limits,so please can anybody help me to solve those?
I have to find a limits for given sequences. Detailed instruction how to solve this would be great. Thank you!

1. \(\displaystyle \lim _{n \to \infty} \frac{2}{3} + \frac{3}{2n^2}\)

2. \(\displaystyle \lim _{n \to \infty} \frac{5n^3+6n-3}{7n-3n^3+2}\)

3. \(\displaystyle \lim _{n \to \infty} n\sqrt{n^2+4}-n^2\)

 

[MarkFL]

Just for future reference, we do ask that no more than 2 questions be asked per thread. This keeps a thread from becoming convoluted and hard to follow.

Let's take these one at a time.

1.) \(\displaystyle \lim _{n\to\infty}\left(\frac{2}{3} + \frac{3}{2n^2} \right)\)

There are 3 theorems I would use here:

a) \(\displaystyle \lim_{x\to c}\left(f(x)\pm g(x) \right)=\lim_{x\to c}f(x)\pm\lim_{x\to c}g(x)\)

b) \(\displaystyle \lim_{x\to c}k=k\) where $k$ is a constant.

c) \(\displaystyle \lim_{x\to\infty}\frac{k}{x^r}=0\) where $0<r$ is a constant.

Can you apply these to get the solution to the given problem?

 

[theakdad]

Just for future reference, we do ask that no more than 2 questions be asked per thread. This keep a thread from becoming convoluted and hard to follow.

Let's take these one at a time.

1.) \(\displaystyle \lim _{n\to\infty}\left(\frac{2}{3} + \frac{3}{2n^2} \right)\)

There are 3 theorems I would use here:

a) \(\displaystyle \lim_{x\to c}\left(f(x)\pm g(x) \right)=\lim_{x\to c}f(x)\pm\lim_{x\to c}g(x)\)

b) \(\displaystyle \lim_{x\to c}k=k\) where $k$ is a constant.

c) \(\displaystyle \lim_{x\to\infty}\frac{k}{x^r}=0\) where $0<r$ is a constant.

Can you apply these to get the solution to the given problem?

\(\displaystyle \lim _{n\to\infty}\frac{2}{3}+ \lim _{n\to\infty}\frac{3}{2n^2}\) ?

= \(\displaystyle \frac{2}{3}+ 0\) ?

 

[MarkFL]

\(\displaystyle \lim _{n\to\infty}\frac{2}{3}+ \lim _{n\to\infty}\frac{3}{2n^2}\) ?

= \(\displaystyle \frac{2}{3}+ 0\) ?

Yes, good! :D

All you need to do is add the two numbers to get the value of the limit.

For the second problem, I recommend dividing each term in both the numerator and denominator by $n^3$, then use the following theorem:

\(\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to c}f(x)}{\lim\limits_{x\to c}g(x)}\)

Are you certain the first term in the numerator has an exponent of 3?

 

[theakdad]

Yes, good! :D

All you need to do is add the two numbers to get the value of the limit.

For the second problem, I recommend dividing each term in both the numerator and denominator by $n^3$, then use the following theorem:

\(\displaystyle \lim_{x\to c}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to c}f(x)}{\lim\limits_{x\to c}g(x)}\)

Are you certain the first term in the numerator has an exponent of 3?

Yes,im sure it has exponent of 3

 

[MarkFL]

Yes,im sure it has exponent of 3

Okay, just wanted to be sure. :D

Allow me to amend my advice to divide each term by $n^2$ instead, as this will make things easier. :D

 

[theakdad]

Okay, just wanted to be sure. :D

Allow me to amend my advice to divide each term by $n^2$ instead, as this will make things easier. :D

Maybe i can do it so:
\(\displaystyle \lim_{x\to \infty} \frac{n^3(5+\frac{6}{n^3}-\frac{3}{n^3})}{n^3(\frac{7n}{n^3}-3+\frac{2}{n^3})}\)

Then i think solution would be: \(\displaystyle \frac{5+0-0}{0-3+0}\) and that's equal to \(\displaystyle - \frac{5}{3}\)

What do you think?

 

[MarkFL]

Maybe i can do it so:
\(\displaystyle \lim_{x\to \infty} \frac{n^3(5+\frac{6}{n^3}-\frac{3}{n^3})}{n^3(\frac{7n}{n^3}-3+\frac{2}{n^3})}\)

Then i think solution would be: \(\displaystyle \frac{5+0-0}{0-3+0}\) and that's equal to \(\displaystyle - \frac{5}{3}\)

What do you think?

You have factored incorrectly. Try dividing each term by $n^2$.

 

[theakdad]

You have factored incorrectly. Try dividing each term by $n^2$.

I think i have factored good...can you look again?
Our teacher said when factoring on limits,we should expose greatest exponent,in this case this is n³.

And when i look to the solution on wolframalpha,i get the same solution.

 

[MarkFL]

You are giving W|A a different problem then. The degree of the numerator (a cubic) is one greater than the degree of the denominator (a quadratic), so it is not going to converge to a finite value.

 

[theakdad]

You are giving W|A a different problem then. The degree of the numerator (a cubic) is one greater than the degree of the denominator (a quadratic), so it is not going to converge to a finite value.

Can u please factor out my first fraction? But with n³ exposed?

I have put in W/A problem as in my first post.

 

[MarkFL]

I see you have now changed the problem so that the denominator is a cubic instead of a quadratic. Dividing each term by $n^3$ we get

\(\displaystyle \lim _{n \to \infty} \frac{5+\dfrac{6}{n^2}-\dfrac{3}{n^3}}{\dfrac{7}{n^2}-3+\dfrac{2}{n^3}}\)

 

[theakdad]

I see you have now changed the problem so that the denominator is a cubic instead of a quadratic. Dividing each term by $n^3$ we get

\(\displaystyle \lim _{n \to \infty} \frac{5+\dfrac{6}{n^2}-\dfrac{3}{n^3}}{\dfrac{7}{n^2}-3+\dfrac{2}{n^3}}\)

Yes ,i have made a mistake,i apologize for that.
But isn't your solution same as mine? Only you have shortened fractions with x³ in them.
So my solution is ok i think,its \(\displaystyle - \frac{5}{3}\) ??

But now I am worried for problem 3,i have no idea how to manage that...

One question about writing in forum? Where do you write the code for "\(\displaystyle ? Do you write it here direct in forum,or have you some kind of latex,and then copy/paste code? Thank you\)

 

[MarkFL]

When you factored, you put an $n^3$ in the denominator of every term that originally had a power on $n$ that was less than 3. You got the right result, but only because those terms go to zero anyway.

To generate the MATH tags, use the button on the toolbar with the $\sum$ character on it.

For the third one, try factoring first and then rationalizing the numerator, as follows:

\(\displaystyle n\sqrt{n^2+4}-n^2=n\left(\sqrt{n^2+4}-n \right)\cdot\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}=?\)

Once you do this then divide each term by $n$.

 

[theakdad]

When you factored, you put an $n^3$ in the denominator of every term that originally had a power on $n$ that was less than 3. You got the right result, but only because those terms go to zero anyway.

To generate the MATH tags, use the button on the toolbar with the $\sum$ character on it.

For the third one, try factoring first and then rationalizing the numerator, as follows:

\(\displaystyle n\sqrt{n^2+4}-n^2=n\left(\sqrt{n^2+4}-n \right)\cdot\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}=?\)

Once you do this then divide each term by $n$.

I jusut thought that there's no need to shorten those fractions,cause i already know that they are 0...important is,,that my factoring was ok,was it?

So the third problem i really don't understand...How did you come t the result you have written?

 

[MarkFL]

I don't know what you mean by shortening fractions...but it is important to factor correctly since it may cause incorrect results with other problems. If you factor out $n^3$, then you must subtract 3 from the exponent of $n$ on each term. If a term is a constant, then it is equivalent to the exponent on $n$ of that term being zero.

For the third problem, rationalizing the numerator and then dividing each term by $n$ as I suggested will give you a determinate form. Check it and see, and you will realize why I suggest it. :D

 

[theakdad]

I don't know what you mean by shortening fractions...but it is important to factor correctly since it may cause incorrect results with other problems. If you factor out $n^3$, then you must subtract 3 from the exponent of $n$ on each term. If a term is a constant, then it is equivalent to the exponent on $n$ of that term being zero.

For the third problem, rationalizing the numerator and then dividing each term by $n$ as I suggested will give you a determinate form. Check it and see, and you will realize why I suggest it. :D

i don't know how to do it...

 

[MarkFL]

Think of the difference of squares formula:

\(\displaystyle a^2-b^2=(a+b)(a-b)\)

 

[theakdad]

Think of the difference of squares formula:

\(\displaystyle a^2-b^2=(a+b)(a-b)\)

i really don't know...

 

[MarkFL]

i really don't know...

In the formula I gave for the difference of squares, let:

\(\displaystyle a=\sqrt{n^2+4},\,b=n\)

So, what would $a^2-b^2$ be?

 

[theakdad]

In the formula I gave for the difference of squares, let:

\(\displaystyle a=\sqrt{n^2+4},\,b=n\)

So, what would $a^2-b^2$ be?

\(\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)\) ??

 

[Petrus]

\(\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)\) ??

let's do something more easy
Can you do this one
\(\displaystyle (3+1)(3-1)\)

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

 

[theakdad]

let's do something more easy
Can you do this one
\(\displaystyle (3+1)(3-1)\)

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

So I am wrong...
Thats \(\displaystyle 9-1\) or 3²- 1²?

 

[MarkFL]

\(\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)\) ??

You have written $(a+b)(a-b)$. You want to write it in its equivalent form $a^2-b^2$.

 

[theakdad]

You have written $(a+b)(a-b)$. You want to write it in its equivalent form $a^2-b^2$.

I think I am confused now,not concentrated

Can you help?

 

[Jameson]

Which one are you stuck on right now? #3 still?

The thing about limits is you'll need to use a handful of other math topics that you've covered before this like exponent rules, factoring, rationalizing part of a fraction by multiplying by the conjugate - those are some I can think of now.

So for #3, why do we need to make it into a fraction in the first place? What should we multiply by to convert the expression into a fraction?

After those parts, you need to simplify $\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)$. Do you remember the FOIL method? You can use that idea here.

 

[Petrus]

So I am wrong...
Thats \(\displaystyle 9-1\) or 3²- 1²?

That is correct!
Look MarkFL post on #20 and do the same thing! I know you can do it just it looks difficult but it is not! trust me!

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

 

[theakdad]

That is correct!
Look MarkFL post on #20 and do the same thing! I know you can do it just it looks difficult but it is not! trust me!

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

\(\displaystyle (\sqrt{n^2+4}+n)^2\) ??

 

[Jameson]

Close. You are looking at this though: $\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)$. There is an $n$ here that you wrote as $n^2$.

Anyway, let's say we want to expand $(a-b)(a+b)$. By FOILing we get $a^2+ab-ab-b^2$. The two terms in the middle cancel each other though so we end up with just $a^2-b^2$. That's where we get $(a-b)(a+b)=a^2-b^2$ from.

In this problem, $a$ and $b$ are not as basic. $a=\sqrt{n^2+4}$ and $b=n$.

$\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)=(a+b)(a-b)=a^2-b^2$ now.

So all you need is to find $a^2$ and $b^2$.

 

[theakdad]

Close. You are looking at this though: $\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)$. There is an $n$ here that you wrote as $n^2$.

Anyway, let's say we want to expand $(a-b)(a+b)$. By FOILing we get $a^2+ab-ab-b^2$. The two terms in the middle cancel each other though so we end up with just $a^2-b^2$. That's where we get $(a-b)(a+b)=a^2-b^2$ from.

In this problem, $a$ and $b$ are not as basic. $a=\sqrt{n^2+4}$ and $b=n$.

$\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)=(a+b)(a-b)=a^2-b^2$ now.

So all you need is to find $a^2$ and $b^2$.

\(\displaystyle (\sqrt{n^2-4})^2 - n^2\) ??

 

[Jameson]

\(\displaystyle \left(\sqrt{n^2-4}\right)^2 - n^2\) ??

Very close again. It should be the $\sqrt{n^2+4}$ with a plus, not a minus. That gives us \(\displaystyle \left(\sqrt{n^2+4}\right)^2 - n^2\)

Now, what happens when we square a square-root? For example, what is \(\displaystyle \left(\sqrt{9} \right)^2\)?

 

[theakdad]

Very close again. It should be the $\sqrt{n^2+4}$ with a plus, not a minus.

Now, what happens when we square a square-root? For example, what is \(\displaystyle \left(\sqrt{9} \right)^2\)?

we can remove square root,so 9 is left

\(\displaystyle n^2+4-n^2\)

 

[Jameson]

we can remove square root,so 9 is left

\(\displaystyle n^2+4-n^2\)

Exactly! That can be simplified even further too. :)

Once you have that you can now write the new fraction that is equivalent to #3. MarkFL showed this process in post #14.

 

[theakdad]

Exactly! That can be simplified even further too. :)

Once you have that you can now write the new fraction that is equivalent to #3. MarkFL showed this process in post #14.

i still don't get it how did i come to here from my first limit \(\displaystyle \lim _{n \to \infty} n\sqrt{n^2+4}-n^2\)

 

[Jameson]

When you factored, you put an $n^3$ in the denominator of every term that originally had a power on $n$ that was less than 3. You got the right result, but only because those terms go to zero anyway.

To generate the MATH tags, use the button on the toolbar with the $\sum$ character on it.

For the third one, try factoring first and then rationalizing the numerator, as follows:

\(\displaystyle n\sqrt{n^2+4}-n^2=n\left(\sqrt{n^2+4}-n \right)\cdot\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}=?\)

Once you do this then divide each term by $n$.

The above is Mark's post as I mentioned. We take the original expression and multiply it by 1 (the top and bottom of the fraction are the same so it's the same as 1). Then we need to simplify a bunch of stuff. That's what we've been doing the last while.

\(\displaystyle n\left(\sqrt{n^2+4}-n \right)\cdot\frac{\sqrt{n^2+4}+n}{\sqrt{n^2+4}+n}=\frac{n\left(\sqrt{n^2+4}-n \right) \left(\sqrt{n^2+4}+n\right)}{\sqrt{n^2+4}+n}\)

The numerator, or top part, of this big fraction has the thing we just simplified so this fraction becomes less complicated now. What is the new fraction?