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[SOLVED] Limit of sequence

wishmaster

Active member
Oct 11, 2013
211
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. \(\displaystyle \lim _{n \to \infty} n\sqrt{n^2+4}-n^2\)
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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?
 

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \lim _{n\to\infty}\frac{2}{3}+ \lim _{n\to\infty}\frac{3}{2n^2}\) ?

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

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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
 

wishmaster

Active member
Oct 11, 2013
211
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 thats equal to \(\displaystyle - \frac{5}{3}\)

What do you think?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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 thats equal to \(\displaystyle - \frac{5}{3}\)

What do you think?
You have factored incorrectly. Try dividing each term by $n^2$.
 

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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.
 

wishmaster

Active member
Oct 11, 2013
211
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

Administrator
Staff member
Feb 24, 2012
13,775
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

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

wishmaster

Active member
Oct 11, 2013
211
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 isnt 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 im 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

Administrator
Staff member
Feb 24, 2012
13,775
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 $n$.
 

wishmaster

Active member
Oct 11, 2013
211
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 theres 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 dont understand...How did you come t the result you have written?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
 

wishmaster

Active member
Oct 11, 2013
211
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 dont know how to do it.....
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Think of the difference of squares formula:

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

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
i realy dont 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?
 

wishmaster

Active member
Oct 11, 2013
211
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

Well-known member
Feb 21, 2013
739
\(\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)\) ??
let's do something more easy
Can you do this one
\(\displaystyle (3+1)(3-1)\)

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

wishmaster

Active member
Oct 11, 2013
211
let's do something more easy
Can you do this one
\(\displaystyle (3+1)(3-1)\)

Regards,
\(\displaystyle |\pi\rangle\)
So im wrong......



Thats \(\displaystyle 9-1\) or 32- 12?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle (\sqrt{n^2+4}+n)(\sqrt{n^2+4}-n)\) ??
You have written $(a+b)(a-b)$. You want to write it in its equivalent form $a^2-b^2$.
 

wishmaster

Active member
Oct 11, 2013
211
You have written $(a+b)(a-b)$. You want to write it in its equivalent form $a^2-b^2$.
I think im confused now,not concentrated

Can you help?
 
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