bstuan

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2 years, 157 days

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These are replies submitted by bstuan

@vv 

Thank you a lot! How can I find rule like this????!!!

@Rouben Rostamian @Axel Vogt:

Thank you a lot for your very detailed explanation. I wonder because I thought, above, you have suggested comparing the functions: f(x) (= orginal function) with g(x) = 1/lnx. Perhaps, @Q190504, the author of the question, also misinterpreted your suggestion, as I did.

@Mariusz Iwaniuk , @Rouben Rostamian : I also don't see a reason to choose g(x) = 1/ln(x).

@Carl Love :

The problem has been resolved.Thank you very much! I also know it is necessary to find another slice that is not a straight line to get a limit that is not 0. However, for inexperienced people like me, it is not easy.

@Carl Love :

With a few variations of the function f(x,y) (as shown: 

choosing the functions gi(x,y) becomes more difficult. The options for choosing g(x,y) with  Ox, Oy axes or y=kx lines all give the result of the limit of f(gi(x,y))=0. Could you suggest me another suitable g(x,y) ? Or another method of proof?

@Carl Love : With your suggestion, the problem is solved. Thanks very much!

@Carl Love 

Sorry for the short typing above. I thought that the limits of g1(x) and g2(x) must always be = 0(?). The difficulty in calculating the limits of limit(x, gi(x)) in this particular problem is that I don't know how to eliminate the indeterminate form, given the gi(x) functions I have chosen.

@Carl Love :

Thank you very much for the detailed instructions, from the basics. But unfortunately, the "easy problem" you left me with is the problem that I am facing. Choosing g1(x) and g2(x) so that their limit = 0 is not difficult. However, calculating the limit of f(g1(x) and f(g2(x) is always difficult because both are "undefined" for the functions gi(x) that I choose. Could you give me an extra suggestion?

@Carl Love : Exactly as you said, because I'm not a native English speaker. I am Vietnamese.

@Carl Love : Thank you a lot again. My mistake.I didn't pay attention to the condition: "exactly when".

@Carl Love: I just read the information you shared. Looks like it's not the case for the problem I'm looking for help with here. In the case of my problem, f(x) (and with it, g(x)) does not converge but diverges.

@Carl Love : Really! It is new for me. Thank you so much.

@Carl Love :

g(x) = 1/x^2 is not appropriate.

Wow! I found it! it is g(x) = 1/x^3. 

@tomleslie :I need to find a function g(x), so that, f(x) >= g(x) >= 0, and that its improper integral is divergent to prove that the improper integral of f(x)too is divergent.

Ah! I understood! My mistake! Sorry for my incorrect word usage.I often use the word "Convergence" to generally refer to the convergence or non-convergence (divergence) of a series or a improper integral.

@tomleslie :Thank you very much! The presentation of results in the style of Maple Tutors is, as such, complicated and less intuitive.

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