Oblique asymptote. Find the limit. Don't use L'Hospital's rule.

[fenilalanin]

I'm trying to find oblique asymptote for [MATH]y=(5x-2)e^{-\frac{1}{3x+1}}[/MATH]. So I need to find the [MATH]\lim_{x \to \infty} (5x-2)e^{-\frac{1}{3x+1} }-5x[/MATH]. The teacher said we are not allowed to use L'Hospital's rule. Please help me (and sorry for my bad english).

 

[Deleted member 4993]

I'm trying to find oblique asymptote for [MATH]y=(5x-2)e^{-\frac{1}{3x+1}}[/MATH]. So I need to find the [MATH]\lim_{x \to \infty} (5x-2)e^{-\frac{1}{3x+1} }-5x[/MATH]. The teacher said we are not allowed to use L'Hospital's rule. Please help me (and sorry for my bad english).

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this assignment.

 

[fenilalanin]

Please show us what you have tried and exactly where you are stuck.

[MATH]f(x)=(5x-2)e^{-\frac{1}{3x+1}}[/MATH][MATH]y=kx+b[/MATH] is the oblique asymptote for [MATH]f(x)[/MATH].
The theorem says [MATH]k=\lim_{x \to \infty}\frac{f(x)}{x}[/MATH] and [MATH]b=\lim_{x \to \infty} f(x)-kx[/MATH]. I don't know any other way to find the oblique asymptote. It would be nice if anyone told me about alternative to this.
In my case, [MATH]k=5[/MATH]. Now I have to find [MATH]b[/MATH]. I know that [MATH]b=\lim_{x \to \infty} (5x-2)e^{-\frac{1}{3x+1} }-5x[/MATH]. The point where I've stuck is solving the limit. It would be easy with L'Hospital's rule, but unfortunately, this rule is prohibited.
Wolfram alpha gave me result [MATH]b=-\frac{11}{3}[/MATH]. But I need detailed solution.
So, the main problem is to find b.

 

[MarkFL]

I would begin by writing the function as:

[MATH]f(x)=\frac{\frac{5}{3}\left(3x+1\right)-\frac{11}{3}}{e^{\frac{1}{3x+1}}}[/MATH]
Let \(u=3x+1\):

[MATH]f(u)=\frac{5u-11}{3e^{\frac{1}{u}}}[/MATH]
[MATH]k=\lim_{u\to\infty}\left(\frac{5u-11}{3ue^{\frac{1}{u}}}\right)=\frac{5}{3}[/MATH]
[MATH]b=\lim_{u\to\infty}\left(\frac{5u-11}{3e^{\frac{1}{u}}}-\frac{5}{3}u\right)=\lim_{u\to\infty}\left(\frac{5u-5ue^{\frac{1}{u}}-11}{3e^{\frac{1}{u}}}\right)[/MATH]
Let's examine the limit:

[MATH]L=\frac{5}{3}\lim_{v\to0}\left(\frac{1-e^v}{ve^v}\right)=\frac{5}{3}\lim_{v\to0}\left(\frac{1-e^v}{v}\right)[/MATH]
Using a Maclaurin series, we may write:

[MATH]1-e^v=1-\sum_{k=0}^{\infty}\left(\frac{v^k}{k!}\right)=-\sum_{k=1}^{\infty}\left(\frac{v^k}{k!}\right)[/MATH]
Divide by \(v\) to get:

[MATH]\frac{1-e^v}{v}=-\sum_{k=1}^{\infty}\left(\frac{v^{k-1}}{k!}\right)=-1-\sum_{k=2}^{\infty}\left(\frac{v^{k-1}}{k!}\right)[/MATH]
Hence:

[MATH]L=-\frac{5}{3}[/MATH]
And so:

[MATH]b=L-\lim_{v\to0}\left(\frac{11}{3e^v}\right)=-\frac{16}{3}[/MATH]
And so the oblique asymptote is:

[MATH]y=\frac{5}{3}u-\frac{16}{3}=\frac{5}{3}(3x+1)-\frac{16}{3}=5x-\frac{11}{3}[/MATH]

 

[Steven G]

I'm trying to find oblique asymptote for [MATH]y=(5x-2)e^{-\frac{1}{3x+1}}[/MATH]. So I need to find the [MATH]\lim_{x \to \infty} (5x-2)e^{-\frac{1}{3x+1} }-5x[/MATH]. The teacher said we are not allowed to use L'Hospital's rule. Please help me (and sorry for my bad english).

I am a bit lost here. Where did the -5x come from?
Isn't the answer just y = 5x-2??

 

[topsquark]

The slant asymptote is [math]y = 5x - \dfrac{11}{3}[/math].




I'm not fond of the method used by the instructor. But the main idea is that we need to find a solution to the limit that is first order in x, which is how you get the slant asymptote. The limit in the exponential (using approximations) contains factors of x that need to be included. There are various ways to do it. MarkFL's is a bit more elegant than mine... I first converted the limit to x = 1/y so I could use a MacLaurin series on the exponential. The details are similar to MarkFL's.

I had decided not to post my solution as it required expansions. I couldn't think of any other way to do it but it seemed that the approximations might be above the OP's level. Is there another way to approach this?

-Dan

 

[fenilalanin]

Thanks everybody for help!

I would begin by writing the function as:

It was a pleasure to read your solution. And I learned some useful tricks. Thank you!

we need to find a solution to the limit that is first order in x

Sorry, I didn't get it. What does "first order in x" mean? I googled this word, and the results were vague. Is it derivative? But when finding the derivative, they divide a small change in f(x) by small change in x, and find the limit. In our case we divided function value by x.
You're right, I don't know anything about expansions. We've studied The Taylor series for the exponential, but we weren't told about The Taylor series in general.

I'm not fond of the method used by the instructor.

To be fair, the instructor doesn't teach us this method for getting the slant asymptote. I read it in a textbook. What about L'Hospital's rule, teacher thinks that it makes solving "algorithmic" and boring.

Also, I found another solution, but I'm not sure.
[MATH](5x-2)e^{-\frac{1}{3x+1}}=5x(e^{-\frac{1}{3x+1}}-1)-2e^{-\frac{1}{3x+1}}[/MATH][MATH]e^x\sim1+x[/MATH] as [MATH]x\to0[/MATH], so in the limit we can replace [MATH](e^{-\frac{1}{3x+1}}-1)[/MATH] with the [MATH]-\frac{1}{3x+1}[/MATH].
[MATH]\lim_{x\to\infty}(-\frac{5x}{3x+1}-2e^{-\frac{1}{3x+1}})[/MATH] is easy.
Am I right?

 

[topsquark]

Sorry, I didn't get it. What does "first order in x" mean?

For example, the MacLaurin expansion (a Taylor series near 0) we get a power series: [math]f(x) = a + bx + cx^2 + \text{ ...}[/math]. A first order approximation is the largest non-constant term in this expansion. Usually this winds up being [math]f(x) \approx a + bx^1[/math], but if b = 0 then we use [math]f(x) \approx a + cx^2[/math]. (Warning: Some people would say that this is a second order approximation because the largest correction term is squared.) So, to first order, [math]e^x \approx 1 + x[/math]. To second order [math]e^x \approx 1 + x + \dfrac{1}{2} x^2[/math], etc.

Also, I found another solution, but I'm not sure.
[MATH](5x-2)e^{-\frac{1}{3x+1}} [B][COLOR=rgb(226, 80, 65)]- 5x[/COLOR] [/B]=5x(e^{-\frac{1}{3x+1}}-1)-2e^{-\frac{1}{3x+1}}[/MATH][MATH]e^x\sim1+x[/MATH] as [MATH]x\to0[/MATH], so in the limit we can replace [MATH](e^{-\frac{1}{3x+1}}-1)[/MATH] with the [MATH]-\frac{1}{3x+1}[/MATH].

Just a minor change.

-Dan