Solve Limits of Sequence: Detailed Instructions

[Jameson]

[math] \left(\sqrt{n^2-4}\right)^2 - n^2[/math] ??

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

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

 

[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 $$\left(\sqrt{9} \right)^2$$?

we can remove square root,so 9 is left

$$n^2+4-n^2$$

 

[Jameson]

we can remove square root,so 9 is left

$$n^2+4-n^2$$

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 [math]\lim _{n \to \infty} n\sqrt{n^2+4}-n^2[/math]

 

[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:

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

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.

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

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?

 

[Jameson]

(Ok, here is my last post on this for now. I think you need to see the next few steps written out so here they are plus some advice)

I think you might want to practice some easier problems first for using these methods. This one has a lot of parts to it if you're seeing it for the first time. Usually limits are given where you have rationalize a fraction somehow and then everything will cancel nicely and easily. After that the cancellation might be trickier. Finally limits like #3 are given and it can be messy.

Being able to quickly choose the proper fraction to multiply with the original limit is key. It is usually going to be the conjugate of the numerator or denominator. Here are some examples.

This is where we are at for #3 so far:

Spoiler

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

From here you need to divide all terms by $n$. That means evaluate

Spoiler

$$\frac{4n}{n}, \frac{\sqrt{n^2+4}}{n},\frac{n}{n}$$

For

Spoiler

$$\frac{\sqrt{n^2+4}}{n}$$ you can use this property: $$\frac{\sqrt{n^2+4}}{n}=\sqrt{\frac{n^2+4}{n^2}} =\sqrt{ \frac{n^2}{n^2}+\frac{4}{n^2}}$$

You hopefully see now why algebra is so important here. Review these concepts some and practice a few easier problems to get a feel for the process. Once you have that you can do this problem in 1-2 minutes probably by knowing the methods you'll probably use.