Limit at infinity with radicals

  • #1

Homework Statement


lim as x tends to -∞ (x)^3/5 - (x)^1/5

Homework Equations




The Attempt at a Solution


The first thing I did was convert it into a radical so it becomes fifthroot√x^3 - fifthroot√x.

Then I rationalized to get ( x^3-x)/(fifthrt√x^3+fifthroot√x) . I then divided the top by x^3 and the bottom by x^3 and x. And the final answer I got was -1/2 .

Im pretty sure I did something wrong so please let me know if my answer is right or wrong.
 

Answers and Replies

  • #2
Simon Bridge
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##(\sqrt[5]{x^3}-\sqrt[5]{x})(\sqrt[5]{x^3}+\sqrt[5]{x}) \neq x^3 - x##
... check by not using the radical notation.
 
  • #3
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If the problem were ##\lim_{x \to -\infty}(x^3 - x)##, the simplest approach would be to factor the expression whose limit is being taken.
 
  • #4
Ray Vickson
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Homework Statement


lim as x tends to -∞ (x)^3/5 - (x)^1/5

Homework Equations




The Attempt at a Solution


The first thing I did was convert it into a radical so it becomes fifthroot√x^3 - fifthroot√x.

Then I rationalized to get ( x^3-x)/(fifthrt√x^3+fifthroot√x) . I then divided the top by x^3 and the bottom by x^3 and x. And the final answer I got was -1/2 .

Im pretty sure I did something wrong so please let me know if my answer is right or wrong.
Your problem might not make sense; the function ##f(x) = x^{3/5}- x^{1/5}## is undefined for ##x< 0##, so we cannot speak assuredly about what happens when ##x \to -\infty##. For ##x = -t## with ##t > 0##, I suppose one can put ##(-t)^{1/5} = - t^{1/5}## (because ##(-1)^5 = -1##) and so re-write the function as ##(-t^{1/5})^3 - (-t^{1/5}) = t^{1/5} - t^{3/5}##, but there are many who would argue against doing that. I don't like it much myself.

Basically, functions like ##z^{3/5}## have so-called principle values in the complex plane, but are essentially "multiple-valued", with a "branch cut" along the negative real axis, which is exactly where you are trying to put your ##z##-values.
 
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  • #5
Simon Bridge
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Yeah - I thought maybe suggesting trying a change of variable x=u^5 and reworking the limit in terms of u... since we have computers, though, the function can simply be plotted to see if it looks like it converges, before trying to get clever with the expression. I suspect there is an important context missing.
 
  • #6
Ray Vickson
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Yeah - I thought maybe suggesting trying a change of variable x=u^5 and reworking the limit in terms of u... since we have computers, though, the function can simply be plotted to see if it looks like it converges, before trying to get clever with the expression. I suspect there is an important context missing.
What about the issue raised in post #4?
 
  • #7
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What about the issue raised in post #4?
From the context of the problem, it seems to me that we're interested in the real-valued function ##x^{3/5} - x^{1/5}##. Both parts are odd roots, so their domain and range is the entire real line. The situation would be different if the roots were even roots, like square roots or fourth roots, and so on.

If you graph either of ##x^{3/5}## or ##x^{1/5}## on wolframalpha, you can specify whether the real-valued root or the principal root.
 
  • #8
Ray Vickson
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From the context of the problem, it seems to me that we're interested in the real-valued function ##x^{3/5} - x^{1/5}##. Both parts are odd roots, so their domain and range is the entire real line. The situation would be different if the roots were even roots, like square roots or fourth roots, and so on.

If you graph either of ##x^{3/5}## or ##x^{1/5}## on wolframalpha, you can specify whether the real-valued root or the principal root.
I know. I did deal with that interpretation in #4, while also saying I did not much like it. I guess my main concern was whether or not the OP even realizes there is an issue. If she does realize there is an issue I would be happier if she mentioned that fact.
 
  • #9
Simon Bridge
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I know. I did deal with that interpretation in #4, while also saying I did not much like it. I guess my main concern was whether or not the OP even realizes there is an issue. If she does realize there is an issue I would be happier if she mentioned that fact.
I agree... you were asking me: "What about the issue raised in post #4?", in response to post #5.
Post #5 is in addition to previous comments, including yours.
The issue is not resolvable without further input from OP... @AlexandraMarie112 ??
 
  • #10
I agree... you were asking me: "What about the issue raised in post #4?", in response to post #5.
Post #5 is in addition to previous comments, including yours.
The issue is not resolvable without further input from OP... @AlexandraMarie112 ??
Sorry everyone! I forgot I asked this question. I figured it out now.
 
  • #11
Simon Bridge
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Sorry everyone! I forgot I asked this question. I figured it out now.
Well done!

You know how frustrating it is when you google for help on a problem, and you find someone with the exact same problem as you, and they say they solved it, and you scroll eagerly for some idea how they solved it, only they don't tell anyone but just say "hey, I figured it out"???
 
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  • #12
Well done!

You know how frustrating it is when you google for help on a problem, and you find someone with the exact same problem as you, and they say they solved it, and you scroll eagerly for some idea how they solved it, only they don't tell anyone but just say "hey, I figured it out"???
LOL. To be honest I just didnt feel like typing but now that you made me realize someone else might need the help , might as well.

Ok well the way I figured out the answer is probably not the way everyone else does. Work:

1) I factored out x^(1/5) . After the factoring Im left with x^(1/5) [ x^(2/5) - (1)]
2) Then with my knowledge of exponents turned it into x^(1/5) [ (x^2)^(1/5) -1]
3) For the x^2 , well I just thought -infty squared is just positive and then just said in my head -∞^(1/5) multiplied by whats in the inner bracket so +∞^(1/5) -1 is just +∞ amd when you multiply these two , the final answer is still -∞

Hope it makes sense to whomever might need the help!
 
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  • #13
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Your problem might not make sense; the function f(x)=x3/5−x1/5 is undefined for x<0
Hi Ray. Sorry, the function is perfectly defined for x<0, just notice that the exponents are 1/5 and 3/5, with the numerator being odd. For example, $$(-3125)^{(3/5)} = -125$$
 
  • #14
Ray Vickson
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Hi Ray. Sorry, the function is perfectly defined for x<0, just notice that the exponents are 1/5 and 3/5, with the numerator being odd. For example, $$(-3125)^{(3/5)} = -125$$
I already said all that in my post #4. At the same time I pointed out that we cannot say assuredly what is the value of something like ##(-3125)^{3/5}##. Of course, the so-called principal value is ##-125## but that is by no means the end of the story.

Here is what results from entering (-3125)^(3/5) and/or (-3125)^.6 into some computer packages.

Maple: (-3125)^(3/5) gives
$$-3125^{3/5} \cos(2\pi/5) +i \, 3125^{3/5} \sin(2 \pi/5) \doteq -38.6271+118.882 \, i $$.
We can also ask Maple to convert the real and imaginary parts to radicals to get
$$(-3125)^{3/5} = \frac{125}{4} - \frac{125 \sqrt{5}}{4} + i\, \frac{125 \sqrt{10 + 2 \sqrt{5}}}{4}$$

(-3125)^.6 gives -38.6271+118.882 i. In all cases we can increase the number of digits in the numerical answer to several thousand digits of precision if we choose.

The simple Maple command "convert((-3125)^(3/5),surd)" produces -125. However, the command "convert((-3125)^.6,surd)" produces the decimal-valued complex number given above.

Wolfram Alpha:
(-3125)^(3/5) gives ##125 (-1)^{3/5} \doteq -38.672 + 118.882 \, i## (but printed out to something like 65 digits of precision), plus other algebraic expressions like the one given above.

Using the command "realvalued root" after getting the output will produce the desired result -125.

EXCEL: (-3125)^(3/5) gives "#NUM!", indicating that EXCEL thinks the answer is not a number.
(-3125)^.6 gives back the printed answer (-2125)^.6. This is seemingly not an error-message, but a simple refusal to do the calculation.

The reason that sophisticated packages like Maple and Mathematica (the driver for Wolfram Alpha) refuse to print -125 right away is that they have built-in checks. They "realize" that taking a fractional power of a negative number is an ambiguous business, and make the user work harder to get a particular answer. For the special case of an odd fractional power they also permit getting the negative real principal value, but they do not just print that out as the official answer.
 
  • #15
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Hi Ray,

Sorry, I didn't realise. In any case I assumed that the OP is working with real numbers.
 
  • #16
epenguin
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Well done!

You know how frustrating it is when you google for help on a problem, and you find someone with the exact same problem as you, and they say they solved it, and you scroll eagerly for some idea how they solved it, only they don't tell anyone but just say "hey, I figured it out"???
Hear, hear. See my sig.

At this level aren't we used to the idea that a negative number has one real negative odd-numbered (third or fifth and so on) root?

We could say here the as we go to big negative x one big negative number is being subtracted from an even bigger negative number, so the result is an ever increasingly big negative number, I.e. This has no limit.

If it makes it any easier we could call x1/5 X; then big negative X is being substracted from the even bigger negative X3..
Alternatively (X3 - X) = X(X2 - 1) so your initial expression is an infinitely big negative number multiplied by name infinitely big positive one, so it is negative and with no limit?
 

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