Finding Asymptotes in Equations: \frac{2x-3}{x-1} and \frac{5(12x+65)}{x^2-25}

[Schrodinger's Dog]

Homework Statement

\frac{2x-3}{x-1} find the asymptotes of this equation.

The Attempt at a Solution

We already know that x=1 is a vertical asymptote from calculating the domain.

the denominator of x-1=0 when x=1.

However I'm not sure what it is they mean when later they go onto say..

f(x)=\frac{2x-3}{x-1}=\frac{2-\frac{3}{x}}{1-\frac{1}{x}}=\frac{2-0}{1-0}=2 as x\rightarrow \infty

What value of x are they using here?

Another more complicated one.

Homework Statement

I don't know what to do to achieve the same say for this:-

\frac {5(12x+65)}{x^2-25}}

What exactly are they doing when they insert the value of x? It doesn't appear they have the same value for x as they do in the numerator as the denominator?

The Attempt at a Solution

This for a graph sketching excercise, by finding certain values you can make a sketch, would I be correct in assuming that there would be a horizontal asymptote in the equation above at y=60, and of course x=-5 and 5 Or am I missing something here?

\frac{60x+\frac{325}{x}}{x^2-\frac{25}{x^2}}=\frac{60}{1}

I think I'm missing something here.

 

[Kurdt]

For the first one they let x tend to infinity and thus the denominator going to infinity makes the fraction tend to zero. That means that there is a non-vertical asymptote at f(x) = 2.

If the numerator is of greater order than the denominator then you'd do a polynomial division and the asymptote would be the leading straight line term.

 

[cristo]

Homework Statement

\frac{2x-3}{x-1} find the asymptotes of this equation.

The Attempt at a Solution

We already know that x=1 is a vertical asymptote from calculating the domain.

the denominator of x-1=0 when x=1.

However I'm not sure what it is they mean when later they go onto say..

f(x)=\frac{2x-3}{x-1}=\frac{2-\frac{3}{x}}{1-\frac{1}{x}}=\frac{2-0}{1-0}=2 as x\rightarrow \infty

What value of x are they using here?

Above, they are just finding the limit of the function as x tends to infinity. This is a standard method of finding limits of quotients of polynomials: divide through by the highest power of x and then use the fact that 1/x tends to zero as x tends to infinity.

Another more complicated one.

Homework Statement

I don't know what to do to achieve the same say for this:-

\frac {5(12x+65)}{x^2-25}}

What exactly are they doing when they insert the value of x? It doesn't appear they have the same value for x as they do in the numerator as the denominator?

The Attempt at a Solution

This for a graph sketching excercise, by finding certain values you can make a sketch, would I be correct in assuming that there would be a horizontal asymptote in the equation above at y=60, and of course x=-5 and 5 Or am I missing something here?

\frac{60x+\frac{325}{x}}{x^2-\frac{25}{x^2}}=\frac{60}{1}

I think I'm missing something here.

I'm not sure what you're doing here.

 

[Kurdt]

The second example you'll have to divide the numerator and denominator by x² then investigate as x tends to plus or minus infinity.

 

[Schrodinger's Dog]

That makes a lot more sense, they don't explain what they have done at all really particularly well, I'll take a look bearing that in mind.

I'm not sure what you're doing here.

Neither was I that was kind of the point, thanks all though.

\frac{\frac{60x}{x^2}+\frac{325}{x^2}}{\frac{x^2}{x^2}-\frac{25}{x^2}}\rightarrow\frac{\frac{60}{x}-0}{1-0}

What would I do now?

I mean I can see \frac {x}{x^2} tends to zero? x\rightarrow\infty

I get it with the simple example but what do I do when I end up with x/x^2?

EDITED: to try the second one, still don't understand what I'm going to end up with here? In the case of 60/x then as x tends to infinity x becomes 0? But then this means the answer is 0? Is that right?

 

[Kurdt]

You will notice that sometimes horizontal asymptotes don't work within the region of vertical asymptotes and the origin and often only apply as x tends to infinity in the positive and negative direction.

 

[Schrodinger's Dog]

Sorry I edited the previous post^^, am I looking a this correctly?

I understand now what they mean but I'm still not sure what the answer would be, is it 60/x? Or do I then go further?

I looked at the graph on Mathcad, there appears to be an asymptote at -5,5, I don't think there exactly is a horizontal asymptote at y=0? There appears to be one at ~y=-4 although I could be imagining it.

 

[Kurdt]

It would of course tend to 0/1 which is just 0. Therefore there is a horizontal asymptote at y=0. As I mentioned before its sometimes not very good around the origin which you can see on your graph and thus there won't be one at y = -4.4. You will of course still get an accurate graph because your investigation of local minimums and maximums will tell you what's going on there.

 

[Schrodinger's Dog]

It would of course tend to 0/1 which is just 0. Therefore there is a horizontal asymptote at y=0. As I mentioned before its sometimes not very good around the origin which you can see on your graph and thus there won't be one at y = -4.4. You will of course still get an accurate graph because your investigation of local minimums and maximums will tell you what's going on there.

Great that's fantastic Kurdt and Cristo, I did a bit of work on this yesterday for the first time, and couldn't really grasp at the time what they were driving at, but had I taken more note I would have seen it, anyway live and learn

 

[Kurdt]

Most textbooks I've seen never explain this concept properly. They just kind of introduce it and never explain that a horizontal asymptote sometimes only works for very large or very small x. And they almost always never explain the method used.

My favourite undergraduate introductory text (Guide 2 Mathematical Methods) which I think is almost perfect in every other way doesn't do this either and it frustrates me immensly.