Question on method of finding limit

[gadav478]

Hello guys-

I have a question about a method of finding the limit as x -> oo fourth root of [(100x+1)/(x^2 + 3x + 2)] (the fourth root of the ENTIRE fraction).
I tried to divide each term by the highest power of x in the denomiator and the professor gave me partial credit for this question, stating "Can't split these roots up". I would have asked him why, but I had to rush to work after class. Can anyone explain why this is?

Also is there a problem I can use in order to display my problem as I write it without all the extra notation? I just want to be clearer next time.

Thanks!

 

[JeffM]

Hello guys-

I have a question about a method of finding the limit as x -> oo fourth root of [(100x+1)/(x^2 + 3x + 2)] (the fourth root of the ENTIRE fraction).
I tried to divide each term by the highest power of x in the denomiator and the professor gave me partial credit for this question, stating "Can't split these roots up". I would have asked him why, but I had to rush to work after class. Can anyone explain why this is?

Also is there a problem I can use in order to display my problem as I write it without all the extra notation? I just want to be clearer next time.

Thanks!

To answer your second question first, this site supports LaTeX, which is a neat way to make mathematical statements render very clearly. Unfortunately, LaTeX takes some time to learn, and unless you find yourself asking many questions, I am not sure it is a productive use of your time. I learned it mostly by hitting "Reply with Quote" to see how people used it, but it took a while and a lot of frustration.

First question. Because you did not show your work, it is impossible to say what you did wrong. We can help you much better if you show your work; please do so in the future even if you know it is wrong.

Another thing that makes it hard to answer is that we do not know what you are allowed to use in your derivations. But I would use (if they are allowed) the general laws of limits. The two that are relevant here are these.

\(\displaystyle \displaystyle \lim_{x \rightarrow a} \left( \sqrt[n]{f(x)}\right) = \sqrt[n]{\lim_{x \rightarrow a} \{f(x) \}}.\)

So find the limit of \(\displaystyle \dfrac{100x + 1}{x^2 + 3x + 2}\) and take the fourth root of the limit.

There is a general set of rules about the limits of rational fractions in x as x goes to plus or minus infinity, but probably those are not yet part of your bag of tricks.

Here is a really important general rule

\(\displaystyle \displaystyle \lim_{x \rightarrow a} \{f(x) * g(x)\} = \left\{\lim_{x \rightarrow a}f(x)\right\} * \left\{\lim_{x \rightarrow a}g(x)\right\}.\)

You can use the rule above in your case as follows

\(\displaystyle \dfrac{100x + 1}{x^2 + 3x + 2} = \dfrac{100x + 1}{(x + 1)(x + 2)} = \dfrac{100x + 1}{x + 1} * \dfrac{1}{x + 2}.\)

Can you take it from there?

 

[DrPhil]

Hello guys-

I have a question about a method of finding the limit as x -> oo fourth root of [(100x+1)/(x^2 + 3x + 2)] (the fourth root of the ENTIRE fraction).
I tried to divide each term by the highest power of x in the denomiator and the professor gave me partial credit for this question, stating "Can't split these roots up". I would have asked him why, but I had to rush to work after class. Can anyone explain why this is?

He obviously wants you to factor the denominator. Do you have the numerator right? if it were 100(x+1) then the (x+1) would cancel. I would have analyzed it just the way you did, but then I'm not a professional mathematician.

Also is there a problem I can use in order to display my problem as I write it without all the extra notation? I just want to be clearer next time.

Are you familiar with LaTeX? There is a good list of symbols at
https://en.wikibooks.org/wiki/LaTeX/Mathematics.
If this string:
\displaystyle \sqrt[4]{\dfrac{100x + 1}{x^2 + 3x + 2}}
is bracketed by tex in square brackets at the beginning and [/tex] at the end, it becomes

\(\displaystyle \displaystyle \sqrt[4]{\dfrac{100x + 1}{x^2 + 3x + 2}} \)

 

[gadav478]

Thanks for the insight. That's a fun looking code lol. I'll give it a whirl.

My work looks like this:

\(\displaystyle \displaystyle \lim_{x \to \infty} \Bigg[ \bigg( \sqrt[4]{\dfrac{100x}{x^2}} + \sqrt[4]{\dfrac{1}{x^2}} \bigg) \bigg/ \bigg( \sqrt[4]{\dfrac{x^2}{x^2}} + \sqrt[4]{\dfrac{3x}{x^2}} + \sqrt[4]{\dfrac{2}{x^2}}\bigg) \Bigg] \)


which I believe implies:

\(\displaystyle \displaystyle ( 0 + 0 ) / (1 + 0 + 0) = 0 \)

I divided each term by the highest \(\displaystyle \displaystyle x \) term to arrive here... is this valid?

 

[DrPhil]

Thanks for the insight. That's a fun looking code lol. I'll give it a whirl.

My work looks like this:

\(\displaystyle \displaystyle \lim_{x \to \infty} \Bigg[ \bigg( \sqrt[4]{\dfrac{100x}{x^2}} + \sqrt[4]{\dfrac{1}{x^2}} \bigg) \bigg/ \bigg( \sqrt[4]{\dfrac{x^2}{x^2}} + \sqrt[4]{\dfrac{3x}{x^2}} + \sqrt[4]{\dfrac{2}{x^2}}\bigg) \Bigg] \)


which I believe implies:

\(\displaystyle \displaystyle ( 0 + 0 ) / (1 + 0 + 0) = 0 \)

I divided each term by the highest \(\displaystyle \displaystyle x \) term to arrive here... is this valid?

Unfortunately, the root of a sum is NOT the sum of the roots!!

That's what he meant by "can't split these roots up like that."

So what you wrote isn't accurate .. the terms must be combined under a single radical. The argument is the same however.

 

[JeffM]

Thanks for the insight. That's a fun looking code lol. I'll give it a whirl.

My work looks like this:

\(\displaystyle \displaystyle \lim_{x \to \infty} \Bigg[ \bigg( \sqrt[4]{\dfrac{100x}{x^2}} + \sqrt[4]{\dfrac{1}{x^2}} \bigg) \bigg/ \bigg( \sqrt[4]{\dfrac{x^2}{x^2}} + \sqrt[4]{\dfrac{3x}{x^2}} + \sqrt[4]{\dfrac{2}{x^2}}\bigg) \Bigg] \)


which I believe implies:

\(\displaystyle \displaystyle ( 0 + 0 ) / (1 + 0 + 0) = 0 \)

I divided each term by the highest \(\displaystyle \displaystyle x \) term to arrive here... is this valid?

You happened to get the right answer, but the method is absolutely wrong.

\(\displaystyle 5 = \sqrt{25} = \sqrt{3^2 + 4^2} \ne \sqrt{3^2} + \sqrt{4^2} = 3 + 4 = 7.\)

You cannot treat roots that way as the example above proves and as your professor said.

I repeat what I said in my first post. The limit of a root (if it exists) is equal to the root of the limit. The roots are distracting you from the real problem, which is finding the limit of the rational fraction if it exists. How to solve the real problem depends on whether DrPhil's guess about what the problem says is correct or not. You need to reread his and my original posts.

If you are still stuck after reading them, please come back and show us what you have tried.

 

[gadav478]

Thanks, I appreciate the help.

By the way, I apologize if it looks as if I ignored your posts, JeffM and DrPhil. I did not. I was just anxious to show my work clearly and use the code so you can SEE what I did. My original question was not concerning a CORRECT method, my question was concerning why the method I used was wrong. In fact, my professor referenced a method for getting the answer totally different than the one you too have referenced.

Again, thanks for the help.

 

[HallsofIvy]

The simplest way to get the limit for this problem is to note that the numerator has degree 1 while the denominator has degree 2. For very large x, this will be approximately \(\displaystyle \sqrt[4]{\frac{1}{x}}\) and it should be obvious that, because the "x" is in the denominator, this goes to 0 as x goes to infinity.

 

[JeffM]

Thanks, I appreciate the help.

By the way, I apologize if it looks as if I ignored your posts, JeffM and DrPhil. I did not. I was just anxious to show my work clearly and use the code so you can SEE what I did. My original question was not concerning a CORRECT method, my question was concerning why the method I used was wrong. In fact, my professor referenced a method for getting the answer totally different than the one you too have referenced.

Again, thanks for the help.

Although I doubt the original poster will be back to look at this thread, his question deserves an answer. To avoid all issues of complex numbers

\(\displaystyle Given\ n \in \mathbb N,\ n \ge 2,\ \sqrt[n]{a} \in \mathbb R,\ \sqrt[n]{b} \in \mathbb R,\ and\ a + b > 0 < ab.\)

\(\displaystyle Let\ c = \sqrt[n]{b * \dfrac{1}{a}} \implies c > 0\ and\ c^n = \dfrac{b}{a} \implies b = ac^n \implies \sqrt[n]{b} = \sqrt[n]{ac^n} = c\sqrt[n]{a} \implies \sqrt[n]{a} + \sqrt[n]{b} = \sqrt[n]{a} + c\sqrt[n]{a} = (1 + c)\sqrt[n]{a}\implies \)

\(\displaystyle \displaystyle \left(\sqrt[n]{a} + \sqrt[n]{b}\right)^n = \left\{(1 + c)\sqrt[n]{a}\right\}^n = a(1 + c)^n = a * \sum_{i=0}^n\left\{\dbinom{n}{i} * 1^{(n-i)} * c^i\right\} = a\left(1^n + c^n +\sum_{i=1}^{n-1}\left\{\dbinom{n}{i} * c^i\right\}\right) = a + b + a\left(\sum_{i=1}^{n-1}\left\{\dbinom{n}{i} * c^i\right\}\right).\)

\(\displaystyle \displaystyle But\ a\left(\sum_{i=1}^{n-1}\left\{\dbinom{n}{i} * c^i\right\}\right) > 0 \implies \left(\sqrt[n]{a} + \sqrt[n]{b}\right)^n \ne a + b \implies \sqrt[n]{a} + \sqrt[n]{b} \ne \sqrt[n]{a + b}.\)

 

[gadav478]

The simplest way to get the limit for this problem is to note that the numerator has degree 1 while the denominator has degree 2. For very large x, this will be approximately \(\displaystyle \sqrt[4]{\frac{1}{x}}\) and it should be obvious that, because the "x" is in the denominator, this goes to 0 as x goes to infinity.

Thanks so much, this make more sense then dividing through by x^2...