How do I solve limits at infinity with similar problems on my exam?

[theakdad]

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 don't get it how to deal with this limits...
I just say thank you for your help!

 

[MarkFL]

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?

 

[theakdad]

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]

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?

 

[theakdad]

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? I am really bad about roots...

 

[MarkFL]

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]

Before trying to do anything, it might be a good idea to rationalise the numerators...

 

[MarkFL]

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)

 

[theakdad]

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 I am 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 don't have a lot of time to learn it,one week actualy! And exam is about induction,sequences and limits.

 

[MarkFL]

\(\displaystyle 4+\frac{1}{n}\) ??

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]

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...

 

[theakdad]

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

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]

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]

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

Is that correct MARK?

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

 

[theakdad]

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]

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]

Can you please show me your way?

No, you can try it yourself first...

 

[theakdad]

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]

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.

 

[theakdad]

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]

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

 

[theakdad]

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]

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

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

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

You made the same error with the radicals as you did with the first problem.

 

[theakdad]

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]

\(\displaystyle \lim _{n \to \infty}\frac{\sqrt{9+0}-2}{\sqrt{9-0}+2}\)

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:

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

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

\(\displaystyle 25=13\)

 

[theakdad]

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:

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

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

\(\displaystyle 25=13\)

thank you!