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[SOLVED] Two limits at infinity

wishmaster

Active member
Oct 11, 2013
211
Again i have two similar problems:

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

\(\displaystyle \lim _{n \to \infty} \frac{\sqrt{9n^2-n}-2n}{\sqrt{9n^2-n}+2n}\)

Those kind of problems will be on my exam,which is very close,and i dont get it how to deal with this limits.....
I just say thank you for your help!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I moved this post into a new thread. We ask that you do not tag new questions onto an existing thread. :D

For both limits, try dividing each term by $n$ to get a determinate form. What do you find?
 

wishmaster

Active member
Oct 11, 2013
211
I moved this post into a new thread. We ask that you do not tag new questions onto an existing thread. :D

For both limits, try dividing each term by $n$ to get a determinate form. What do you find?
I apologize for the thread.......

\(\displaystyle \lim _{n \to \infty} \frac{n(3-\sqrt{4n+1}}{n(3+\sqrt{4n-1}}\)

\(\displaystyle \lim _{n \to \infty} \frac{n(\sqrt{9n-1}+2}{n(\sqrt{9n+1}-2}\)

Is that ok?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I apologize for the thread.......

\(\displaystyle \lim _{n \to \infty} \frac{n(3-\sqrt{4n+1}}{n(3+\sqrt{4n-1}}\)

\(\displaystyle \lim _{n \to \infty} \frac{n(\sqrt{9n-1}+2}{n(\sqrt{9n+1}-2}\)

Is that ok?
That's not correct. When factoring out an $n$, you need to factor out an $n^2$ from the radicals.

Let's look at the first one:

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

Now, dividing each term by $n$ we obtain:

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

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

Now, can you simplify this?
 

wishmaster

Active member
Oct 11, 2013
211
That's not correct. When factoring out an $n$, you need to factor out an $n^2$ from the radicals.

Let's look at the first one:

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

Now, dividing each term by $n$ we obtain:

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

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

Now, can you simplify this?
Can you give me a hint? Im really bad about roots.......
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's just look at the expression under the radical in the numerator:

\(\displaystyle \frac{4n^2+n}{n^2}=\frac{4n^2}{n^2}+\frac{n}{n^2}\)

What is the next step to simplify that?
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Before trying to do anything, it might be a good idea to rationalise the numerators...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Before trying to do anything, it might be a good idea to rationalise the numerators...
Don't you think in the case of the two posted problems, this unnecessarily complicates matters? (Emo)
 

wishmaster

Active member
Oct 11, 2013
211
Let's just look at the expression under the radical in the numerator:

\(\displaystyle \frac{4n^2+n}{n^2}=\frac{4n^2}{n^2}+\frac{n}{n^2}\)

What is the next step to simplify that?
\(\displaystyle 4+\frac{1}{n}\) ??

I know im not good with roots,any idea how can i improve this skill? Any good web site?
When i was younger,that was not a problem,but as i get older,seems my brain gets slower.....and i dont have a lot of time to learn it,one week actualy! And exam is about induction,sequences and limits.
 

MarkFL

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Staff member
Feb 24, 2012
13,775
\(\displaystyle 4+\frac{1}{n}\) ??
Yes, that's correct! (Sun)

Now, can you fully simplify:

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

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Don't you think in the case of the two posted problems, this unnecessarily complicates matters? (Emo)
Not at all, in fact, it eliminates nearly all the square roots, making the simplification easier...
 

wishmaster

Active member
Oct 11, 2013
211
\(\displaystyle \lim_{n\to\infty}\frac{3-\sqrt{4+\frac{1}{n}}}{3+\sqrt{4-\frac{1}{n}}}\)

Is that correct MARK?

- - - Updated - - -

Not at all, in fact, it eliminates nearly all the square roots, making the simplification easier...

Can you please show me your way?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Not at all, in fact, it eliminates nearly all the square roots, making the simplification easier...
I would agree that the number of radicals is reduced by one for the second problem since the radicands are the same, but they are different in the first problem. I still think dividing each term by $n$ in both problems is the simplest way to go computationally.

But...what one person finds simpler may not be the same for another. :D
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \lim_{n\to\infty}\frac{3-\sqrt{4+\frac{1}{n}}}{3+\sqrt{4-\frac{1}{n}}}\)

Is that correct MARK?
Yes...now what happens to $\dfrac{1}{n}$ as $n\to\infty$?
 

wishmaster

Active member
Oct 11, 2013
211
Yes...now what happens to $\dfrac{1}{n}$ as $n\to\infty$?
\(\displaystyle \frac{1}{n}\) is zero.

So i think my solution is \(\displaystyle \frac{3-2+0}{3+2-0} = \frac{1}{5}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It would be better for you to write:

\(\displaystyle \lim_{n\to\infty}\frac{3-\sqrt{4+\dfrac{1}{n}}}{3+\sqrt{4-\dfrac{1}{n}}}=\frac{3-\sqrt{4+0}}{3+\sqrt{4-0}}=\frac{3-\sqrt{4}}{3+\sqrt{4}}=\frac{3-2}{3+2}=\frac{1}{5}\)

The way you wrote it implies:

\(\displaystyle \sqrt{a+b}=\sqrt{a}+\sqrt{b}\)

which is not true in general.
 

Prove It

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MHB Math Helper
Jan 26, 2012
1,404

wishmaster

Active member
Oct 11, 2013
211
It would be better for you to write:

\(\displaystyle \lim_{n\to\infty}\frac{3-\sqrt{4+\dfrac{1}{n}}}{3+\sqrt{4-\dfrac{1}{n}}}=\frac{3-\sqrt{4+0}}{3+\sqrt{4-0}}=\frac{3-\sqrt{4}}{3+\sqrt{4}}=\frac{3-2}{3+2}=\frac{1}{5}\)

The way you wrote it implies:

\(\displaystyle \sqrt{a+b}=\sqrt{a}+\sqrt{b}\)

which is not true in general.
Thank you! I think with some practice,i would get the point.....
Any ideas where could i find similar problems to solve?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Thank you! I think with some practice,i would get the point.....
Any ideas where could i find similar problems to solve?
Can you now work the second problem in the same way?

I personally don't know of any online collections of such problems.
 

wishmaster

Active member
Oct 11, 2013
211
Can you now work the second problem in the same way?

I personally don't know of any online collections of such problems.
Any other idea how could i improve?

Ok,the second problem:

\(\displaystyle \lim _{n \to \infty} \frac{\sqrt{9n^2-n}-2n}{\sqrt{9n^2-n}+2n}\)

=\(\displaystyle \lim _{n \to \infty} \frac{\sqrt{\frac{9n^2-n}{n^2}}-\frac{2}{n}}{\sqrt{\frac{9n^2-n}{n^2}}+\frac{2}{n}}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Any other idea how could i improve?

Ok,the second problem:

\(\displaystyle \lim _{n \to \infty} \frac{\sqrt{9n^2-n}-2n}{\sqrt{9n^2-n}+2n}\)

=\(\displaystyle \lim _{n \to \infty} \frac{\sqrt{\frac{9n^2-n}{n^2}}-\frac{2}{n}}{\sqrt{\frac{9n^2-n}{n^2}}+\frac{2}{n}}\)
The expressions under the radicals are correct, but the other terms should be:

\(\displaystyle \frac{2n}{n}\)

Once you correct this, then continue to simplify like you did in the first problem. :D
 

wishmaster

Active member
Oct 11, 2013
211
The expressions under the radicals are correct, but the other terms should be:

\(\displaystyle \frac{2n}{n}\)

Once you correct this, then continue to simplify like you did in the first problem. :D
\(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9+\frac{1}{n}}-2}{\sqrt{9-\frac{1}{n}}+2}\)

= \(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9}+\sqrt{\frac{1}{n}}-2}{\sqrt{9}-\sqrt{\frac{1}{n}}+2}\)

= \(\displaystyle \frac{3+0-2}{3-0+2}\) = \(\displaystyle \frac{1}{5}\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9+\frac{1}{n}}-2}{\sqrt{9-\frac{1}{n}}+2}\)

= \(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9}+\sqrt{\frac{1}{n}}-2}{\sqrt{9}-\sqrt{\frac{1}{n}}+2}\)

= \(\displaystyle \frac{3+0-2}{3-0+2}\) = \(\displaystyle \frac{1}{5}\)
You made the same error with the radicals as you did with the first problem.
 

wishmaster

Active member
Oct 11, 2013
211
You made the same error with the radicals as you did with the first problem.
\(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9+0}-2}{\sqrt{9-0}+2}\)

Is that ok now?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9+0}-2}{\sqrt{9-0}+2}\)

Is that ok now?
Yes. While what you did did not affect the final answer in these problems, this kind of algebraic error could get you into trouble on down the line with other problems. It is a form of an error so commonly made it is referred to as "The Freshman's Dream" in which many many students try to use:

\(\displaystyle (a+b)^n=a^n+b^n\)

And this is wrong. Consider:

\(\displaystyle (2+3)^2=2^2+3^2\)

\(\displaystyle 5^2=4+9\)

\(\displaystyle 25=13\)