Limits of sequences

[jpanknin]

This approach isn't specific to sequences, but that's where it came up in the textbook. The following example divides both the numerator and the denominator by n, the highest power of n in the denominator (this technique is explained in the "Solution" section).




So both are divided by n. No problem. That makes sense to me. But in the following example, the solution divides both the numerator and denominator by [imath]n^2[/imath]. But isn't the highest power of n in the denominator [imath]\sqrt{n}[/imath] instead of [imath]n^2[/imath]? The "Solution" for example 4 shown above says divide by the highest power of n that occurs in the denominator. This seems to depart from that.



So why is it [imath]n^2[/imath] in Example 5?

 

[fresh_42]

Because [imath] a\cdot \sqrt{b} = \sqrt{a^2\cdot b}[/imath] or the other way round [imath] \sqrt{a^2\cdot b}=\sqrt{a^2}\cdot \sqrt{b}=a\sqrt{b}. [/imath]

 

[jpanknin]

Still not following. If the denominator is [imath]\sqrt{10 + n}[/imath], then isn't the HIGHEST power of n in the denominator [imath]\sqrt{n}[/imath] rather than n? It seems the highest power in the denominator is 1/2 rather than 1.

 

[lev888]

So why is it [imath]n^2[/imath] in Example 5?

Because when n is admitted inside the square root it's granted the power of 2.

 

[lev888]

Still not following. If the denominator is [imath]\sqrt{10 + n}[/imath], then isn't the HIGHEST power of n in the denominator [imath]\sqrt{n}[/imath] rather than n? It seems the highest power in the denominator is 1/2 rather than 1.

What do you get if you divide everything by [imath]\sqrt{n}[/imath]?

 

[Dr.Peterson]

So why is it [imath]n^2[/imath] in Example 5?

You divide numerator and denominator each by n; it is only [imath]n^2[/imath] when moved inside the radical.

When I see an expression like that denominator, I imagine the lower power (the 10) going away (because it is relatively small, for large n); what's left is [imath]\sqrt{n}[/imath]. That's the highest power in the denominator. Think of it as "the highest power after processing", not "as is".

So the highest power in the entire expression (both numerator and denominator), is just [imath]n[/imath]; and when we divide the denominator by [imath]n[/imath], we have to move it inside the radical by squaring it.

Still not following. If the denominator is [imath]\sqrt{10 + n}[/imath], then isn't the HIGHEST power of n in the denominator [imath]\sqrt{n}[/imath] rather than n? It seems the highest power in the denominator is 1/2 rather than 1.

Yes, that highest power is [imath]\sqrt{n}[/imath] (i.e., exponent 1/2). But they divide top and bottom by the same thing, which is the highest power across both parts.

So what they're doing (ignoring the limit) is dividing by [imath]n[/imath]: [math]\frac{n}{\sqrt{10+n}}=\frac{\frac{n}{n}}{\frac{\sqrt{10+n}}{n}}=\frac{1}{\sqrt{\frac{10+n}{n^2}}}=\frac{1}{\sqrt{\frac{10}{n^2}+\frac{n}{n^2}}}=\frac{1}{\sqrt{\frac{10}{n^2}+\frac{1}{n}}}[/math]

 

[jpanknin]

You divide numerator and denominator each by n; it is only [imath]n^2[/imath] when moved inside the radical.

When I see an expression like that denominator, I imagine the lower power (the 10) going away (because it is relatively small, for large n); what's left is [imath]\sqrt{n}[/imath]. That's the highest power in the denominator. Think of it as "the highest power after processing", not "as is".

So the highest power in the entire expression (both numerator and denominator), is just [imath]n[/imath]; and when we divide the denominator by [imath]n[/imath], we have to move it inside the radical by squaring it.


Yes, that highest power is [imath]\sqrt{n}[/imath] (i.e., exponent 1/2). But they divide top and bottom by the same thing, which is the highest power across both parts.

So what they're doing (ignoring the limit) is dividing by [imath]n[/imath]: [math]\frac{n}{\sqrt{10+n}}=\frac{\frac{n}{n}}{\frac{\sqrt{10+n}}{n}}=\frac{1}{\sqrt{\frac{10+n}{n^2}}}=\frac{1}{\sqrt{\frac{10}{n^2}+\frac{n}{n^2}}}=\frac{1}{\sqrt{\frac{10}{n^2}+\frac{1}{n}}}[/math]

So you're saying you should take the highest power of EITHER the numerator or denominator? Am I understanding that correctly? Because the guidance in the book specifically said the DENOMINATOR.

Also, in a later problem (see below), they did use [imath]\sqrt{n^3}[/imath]. Following the logic of the original problem above, I used [imath]\sqrt{n^6}[/imath], but that was wrong (see second pic below and sorry for the 3rd grade handwriting...). I think I followed the same logic.

 

[lookagain]

View attachment 38399

For rewriting \(\displaystyle \ a_n, \ \) multiply numerator and denominator by \(\displaystyle \ \tfrac{1}{n^2} \ \) instead. Then, you won't end up with a 0/0 form.

 

[mario99]

Don't use division approach. It is confusing to new students. Instead use common factor approach.


[imath]\displaystyle n^3 + 4n = \frac{n^4}{n} + \frac{4n^4}{n^3} = n^4\left(\frac{1}{n} + \frac{4}{n^3}\right)[/imath]


[imath]\displaystyle \lim_{n \rightarrow \infty}\frac{n^2}{\sqrt{n^3 + 4n}} = \lim_{n \rightarrow \infty}\frac{n^2}{\sqrt{n^4\left(\frac{1}{n} + \frac{4}{n^3}\right)}} = \lim_{n \rightarrow \infty}\frac{n^2}{n^2\sqrt{\frac{1}{n} + \frac{4}{n^3}}}[/imath]


[imath]\displaystyle = \lim_{n \rightarrow \infty}\frac{1}{\sqrt{\frac{1}{n} + \frac{4}{n^3}}} = \frac{1}{\sqrt{0 + 0}} = \frac{1}{\sqrt{0}} = \frac{1}{0} = \infty[/imath]

 

[lookagain]

For rewriting \(\displaystyle \ a_n, \ \) multiply numerator and denominator by \(\displaystyle \ \tfrac{1}{n^2} \ \) instead. Then, you won't end up with a 0/0 form.

jpanknin, you should be able to handle what I stated, combined with your attempt in
post # 7.

Post # 9 does not specifically address you by username. It presumes that you
are confused by the division approach as a new student and are not to use it.

I don't presume you should not use it, but that you should be able to do it.
It's not your fault that the author did not put enough connecting steps in the
solution to example 5.

 

[mario99]

Post # 9 does not specifically address you by username. It presumes that you
are confused by the division approach as a new student and are not to use it.

I don't presume you should not use it, but that you should be able to do it.

Professor lookagain, thanks for your interest in my post. My name is Mario. And I was a student once upon a time.

When I solved the limits for the first time, I got in the same shoes where the OP is now. I discovered that the division approach when was used with roots in the denominator, a wrong result most likely happened. In the other hand, I have never got a wrong result by using the common factor approach.

I am only sharing my experience. And the OP has the choice to choose any method. But he will have to understand the division approach pretty good if that what his instructor wants him to use in the test.

 

[Dr.Peterson]

So you're saying you should take the highest power of EITHER the numerator or denominator? Am I understanding that correctly? Because the guidance in the book specifically said the DENOMINATOR.

I didn't say you should, just that they did.

Also, in a later problem (see below), they did use [imath]\sqrt{n^3}[/imath]. Following the logic of the original problem above, I used [imath]\sqrt{n^6}[/imath], but that was wrong (see second pic below and sorry for the 3rd grade handwriting...). I think I followed the same logic.

View attachment 38397

View attachment 38399

I have no idea why they are inconsistent; perhaps if I read the whole chapter, I would see that they have secretly explained it. Often either way works.

You just have to do it correctly. (In this example, the highest power overall is [imath]n^2[/imath], not [imath]n^3[/imath]; and as you've been told, that, too, works -- just differently.