Trouble with calculating oblique asymptote (y=mx+q) for f(x)=(x^(2/5))(5-x)^(3/5)

[Ale_ygt]

Hi, I have to calculate the oblique asymptote (y = mx + q) for
f(x) = (x^2/5)(5-x)^3/5
The slope was trivial to find( lim_x->∞[f(x)/x]= -1 = m ), the y-intercept is what's giving me quite some trouble(or in other words, calculating lim_x->∞[f(x) - mx] = lim_x->∞[(x^2/5)(5-x)^3/5 + x] ).

I went as far as converting it into the indeterminate form 0/0(at least I think it is 0/0): lim_x->∞ { [(5-x)^3/5 + x^3/5] / x^-2/5 } and tried to apply L'Hopital's rule, but after 2 iterations it didn't seem like I was going anywhere :???:

Is there anything I got wrong/any method I'm unaware of to solve it? Theoretically the result should be 3 (so that the asymptote is y = -x+3).
Thanks in advance

 

[tkhunny]

Hi, I have to calculate the oblique asymptote (y = mx + q) for
f(x) = (x^2/5)(5-x)^3/5
The slope was trivial to find( lim_x->∞[f(x)/x]= -1 = m ), the y-intercept is what's giving me quite some trouble(or in other words, calculating lim_x->∞[f(x) - mx] = lim_x->∞[(x^2/5)(5-x)^3/5 + x] ).

I went as far as converting it into the indeterminate form 0/0(at least I think it is 0/0): lim_x->∞ { [(5-x)^3/5 + x^3/5] / x^-2/5 } and tried to apply L'Hopital's rule, but after 2 iterations it didn't seem like I was going anywhere :???:

Is there anything I got wrong/any method I'm unaware of to solve it? Theoretically the result should be 3 (so that the asymptote is y = -x+3).
Thanks in advance

1) Why do you think there is an Oblique Asymptote?
2) Are you SURE there is only ONE?

 

[mmm4444bot]

… quite some trouble … calculating …

lim_x->∞[(x^2/5)(5-x)^3/5 + x]

… the result should be 3 (so that the asymptote is y=-x+3) …

If function f uses real-valued roots, I agree.

If you're dealing with principal roots, instead, then you'd better check the function's domain.

I asked Wolfram for a hint on evaluating that limit, and it told me to "rationalize the expression" first. (That hint didn't help me, so we'll both need to wait for help.)

Here's a graph of f(x) -- using real-valued roots -- and its asymptote.

 

[Ale_ygt]

1) Why do you think there is an Oblique Asymptote?

I checked the function's graph(see mmm4444bot's attachment) and it looks like there is an oblique asymptote by all accounts... though it's mostly an observation, the only way to know for sure would be solving the limit I described in the original post.

2) Are you SURE there is only ONE?

Well, f(x) is defined for all real numbers so there can't be any vertical asymptotes/discontinuities, and f(x) tends to infinity as x tends to infinity so there can't be any horizontal asymptotes either; so yeah, I'd say there is just one.

If function f uses real-valued roots, I agree.

If you're dealing with principal roots, instead, then you'd better check the function's domain.

Not quite sure about the difference between real-valued and principal roots, if real-valued means only the real solutions without complex numbers then yeah, that's the case.

 

[Dr.Peterson]

Hi, I have to calculate the oblique asymptote (y = mx + q) for
f(x) = (x^2/5)(5-x)^3/5
The slope was trivial to find( lim_x->∞[f(x)/x]= -1 = m ), the y-intercept is what's giving me quite some trouble(or in other words, calculating lim_x->∞[f(x) - mx] = lim_x->∞[(x^2/5)(5-x)^3/5 + x] ).

I went as far as converting it into the indeterminate form 0/0(at least I think it is 0/0): lim_x->∞ { [(5-x)^3/5 + x^3/5] / x ^-2/5 } and tried to apply L'Hopital's rule, but after 2 iterations it didn't seem like I was going anywhere :???:

Is there anything I got wrong/any method I'm unaware of to solve it? Theoretically the result should be 3 (so that the asymptote is y = -x+3).
Thanks in advance

I asked Wolfram for a hint on evaluating that limit, and it told me to "rationalize the expression" first. (That hint didn't help me, so we'll both need to wait for help.)

I think what they are talking about is multiplying the numerator and denominator by something that will eliminate the 5th roots. The way to do this is to use the fact that (a + b)(a⁴ - a³b + a²b² - ab³ + b⁴) = a⁵ + b⁵. Take a = (5-x)^3/5 and b = x^3/5, and multiply the numerator and denominator of your fraction by (a⁴ - a³b + a²b² - ab³ + b⁴). You will find (with some twisting of your mind) that you can "cancel" the highest remaining power, x², from numerator and denominator (that is, multiply both by x^-2), and you will be left with a fraction in which you can take x to infinity and get 3 for the answer. It would be too much work to write it all out for you, and it's better for you if you work out the details yourself!

I would think that L'Hopital would be easier, but I haven't tried that direction yet. Probably we will have to make a twist somewhere, changing the form of the expression so that we aren't just doing the same thing every time.

 

[Ale_ygt]

Alright, I did as suggested by Dr. Peterson and obtained this (I suggest pasting the formulas here for a more comprehensible formatting):
(x^(2/5)((5-x)^3 + x^3)) / ( (5-x)^(12/5) - (5-x)^(9/5)*x^(3/5) + (5-x)^(6/5)*x^(6/5) - (5-x)^(3/5)*x^(9/5) + x^(12/5))

Then I divided numerator and denominator by x^2/5 and obtained this:
(((5-x)^3 + x^3)) / ( ((5-x)^(12/5))/x^(2/5) - (5-x)^(9/5)*x^(1/5) + (5-x)^(6/5)*x^(4/5) - (5-x)^(3/5)*x^(7/5) + x^(2))

...and then I got stuck again
I was thinking about dividing both numerator and denominator by either x³ or (5-x)³ as the next step, but it looks like the denominator gets even more messed up, and applying L'Hopital's rule makes the denominator even worse; unless I'm missing something else, I'm afraid it's just one of those problems with no easy solutions, involving lots of steps and patience, but at this point I guess I'll just accept the fact that the result is 3 and move on.
Thanks everyone

 

[Dr.Peterson]

Alright, I did as suggested by Dr. Peterson and obtained this (I suggest pasting the formulas here for a more comprehensible formatting):
(x^(2/5)((5-x)^3 + x^3)) / ( (5-x)^(12/5) - (5-x)^(9/5)*x^(3/5) + (5-x)^(6/5)*x^(6/5) - (5-x)^(3/5)*x^(9/5) + x^(12/5))

Then I divided numerator and denominator by x^2/5 and obtained this:
(((5-x)^3 + x^3)) / ( ((5-x)^(12/5))/x^(2/5) - (5-x)^(9/5)*x^(1/5) + (5-x)^(6/5)*x^(4/5) - (5-x)^(3/5)*x^(7/5) + x^(2))

...and then I got stuck again
I was thinking about dividing both numerator and denominator by either x³ or (5-x)³ as the next step, but it looks like the denominator gets even more messed up, and applying L'Hopital's rule makes the denominator even worse; unless I'm missing something else, I'm afraid it's just one of those problems with no easy solutions, involving lots of steps and patience, but at this point I guess I'll just accept the fact that the result is 3 and move on.
Thanks everyone

Yes, I said you'd need a little twist!

You may recall from limits of rational functions that dividing everything by the highest power, so that most powers are negative, can help with limits at infinity. That's the purpose of my suggestion to multiply by x^-2 in numerator and denominator.

First, expand the numerator; you'll find that x^3 cancels out, leaving x^2 as the highest power. To get rid of that (so that the numerator will be a finite number when x goes to infinity), we multiply by x^-2. When we do that in the denominator, keeping our goal in mind, we can do this (looking only at the first term:

((5-x)^(12/5))/x^(2/5) * x^-2) = (5-x)^(12/5) * x^(-2/5) * x^-2 = (5-x)^(12/5) * x^(-12/5) = ((5-x)*x^-1)^(12/5) = (5/x - 1)^(12/5)

In the limit, this term will go to 1. Do the same with everything else, and you'll get to your desired result.

 

[Ale_ygt]

At long last I'd say I got it
Continuing from where I left off in the previous post:


Expand the numerator:
\(\displaystyle
\frac{5((5-x)^2 -(5-x)x + x^2)} {\frac{(5-x)^{12/5}}{x^{2/5}}-(5-x)^{9/5}*x^{1/5} + (5-x)^{6/5}*x^{4/5} - (5-x)^{3/5}*x^{7/5} + x^2}
=\frac{5((x^2)(\frac{5}{x}- 1)^2 -x^2(\frac{5}{x}-1) + x^2)}{\frac{(5-x)^{12/5}}{x^{2/5}}-(5-x)^{9/5}*x^{1/5} + (5-x)^{6/5}*x^{4/5} - (5-x)^{3/5}*x^{7/5} + x^2}
\)


divide numerator and denominator by x^2
\(\displaystyle =\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ \frac{(5-x)^{12/5}}{x^{12/5}} - \frac{(5-x)^{9/5}}{x^{9/5}} + \frac{(5-x)^{6/5}}{x^{6/5}} - \frac{(5-x)^{3/5}}{x^{3/5}} + 1}}\)

Some modifications to compute the limit on the denominator in an easier way:
\(\displaystyle
=\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ \frac{x^{12/5}(\frac{5}{x}-1)^{12/5}}{x^{12/5}} - \frac{x^{9/5}(\frac{5}{x}-1)^{9/5}}{x^{9/5}} + \frac{x^{6/5}(\frac{5}{x}-1)^{6/5}}{x^{6/5}} - \frac{x^{3/5}(\frac{5}{x}-1)^{3/5}}{x^{3/5}} + 1}}

=\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ (\frac{5}{x}-1)^{12/5} - (\frac{5}{x}-1)^{9/5} + (\frac{5}{x}-1)^{6/5} - (\frac{5}{x}-1)^{3/5} + 1}}
\)

Time to compute the limit:
\(\displaystyle
\lim_{x\to\infty}\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ (\frac{5}{x}-1)^{12/5} - (\frac{5}{x}-1)^{9/5} + (\frac{5}{x}-1)^{6/5} - (\frac{5}{x}-1)^{3/5} + 1}}

=\frac{5(3)}{1 - (-1) + 1 - (-1) + 1}

=\frac{5(3)}{5}= 3
\)


Huge thanks to Dr. Peterson for all the support, much appreciated

 

[Dr.Peterson]

At long last I'd say I got it
Continuing from where I left off in the previous post:


Expand the numerator:
\(\displaystyle
\frac{5((5-x)^2 -(5-x)x + x^2)} {\frac{(5-x)^{12/5}}{x^{2/5}}-(5-x)^{9/5}*x^{1/5} + (5-x)^{6/5}*x^{4/5} - (5-x)^{3/5}*x^{7/5} + x^2}
=\frac{5((x^2)(\frac{5}{x}- 1)^2 -x^2(\frac{5}{x}-1) + x^2)}{\frac{(5-x)^{12/5}}{x^{2/5}}-(5-x)^{9/5}*x^{1/5} + (5-x)^{6/5}*x^{4/5} - (5-x)^{3/5}*x^{7/5} + x^2}
\)


divide numerator and denominator by x^2
\(\displaystyle =\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ \frac{(5-x)^{12/5}}{x^{12/5}} - \frac{(5-x)^{9/5}}{x^{9/5}} + \frac{(5-x)^{6/5}}{x^{6/5}} - \frac{(5-x)^{3/5}}{x^{3/5}} + 1}}\)

Some modifications to compute the limit on the denominator in an easier way:
\(\displaystyle
=\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ \frac{x^{12/5}(\frac{5}{x}-1)^{12/5}}{x^{12/5}} - \frac{x^{9/5}(\frac{5}{x}-1)^{9/5}}{x^{9/5}} + \frac{x^{6/5}(\frac{5}{x}-1)^{6/5}}{x^{6/5}} - \frac{x^{3/5}(\frac{5}{x}-1)^{3/5}}{x^{3/5}} + 1}}

=\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ (\frac{5}{x}-1)^{12/5} - (\frac{5}{x}-1)^{9/5} + (\frac{5}{x}-1)^{6/5} - (\frac{5}{x}-1)^{3/5} + 1}}
\)

Time to compute the limit:
\(\displaystyle
\lim_{x\to\infty}\frac{5((\frac{5}{x} - 1)^2 -(\frac{5}{x}-1) + 1)}{{ (\frac{5}{x}-1)^{12/5} - (\frac{5}{x}-1)^{9/5} + (\frac{5}{x}-1)^{6/5} - (\frac{5}{x}-1)^{3/5} + 1}}

=\frac{5(3)}{1 - (-1) + 1 - (-1) + 1}

=\frac{5(3)}{5}= 3
\)


Huge thanks to Dr. Peterson for all the support, much appreciated

You did things in a very different order than I did, but it looks like you did everything that was needed!

When we say to "expand" an expression, we usually mean to distribute (eliminating parentheses). In the numerator you had (5-x)^3 + x^3), which expands to 125 - 75x + 15x^2 - x^3 + x^3 = 125 - 75x + 15x^2, which is where I said the highest degree was 2. What you did skipped past that, but worked fine.

 

[Ale_ygt]

Ah, my bad; I always called what I did in the numerator "expanding a sum of cubes", but a quick research reveals the proper nomenclature is factoring a sum of cubes. If I took the "proper expansion" route the solution would have certainly been more elegant, but what I did luckily seemed to work anyway