# Linear oblique asymptote

1. Dec 4, 2013

### Coco12

1. The problem statement, all variables and given/known data

what is the linear oblique asymptote of (x^5+x^3+2)/(x^4-1)
?
2. Relevant equations

x-a/p(x) = q(x) +remainder

3. The attempt at a solution

I put in all the placeholders for the divisor and the numerator and got x as the equation for the linear oblique asymptote?? Is that right??

2. Dec 4, 2013

### Mentallic

We would let

$$\frac{x^5+x^3+2}{x^4-1}\equiv \frac{(ax+b)(x^4-1)+p(x)}{x^4-1} = ax+b + \frac{p(x)}{x^4-1}$$

Where p(x) is a cubic polynomial or less (doesn't matter what it is exactly).

If we expanded (ax+b)(x4-1) then we get

$$ax^5+bx^4-ax-b$$

But we ignore the -ax-b term because that will be a part of p(x) which we've already said we don't care about. So we want the constant a to be chosen such that $ax^5=x^5$ since the coefficient of $x^5$ on the LHS must be equal to the RHS, hence a=1, and b must be chosen such that $bx^4=0$ for the same reason, hence b=0.

But we ignore the -ax-b term because that will be a part of p(x) which we've already said we don't care about.

3. Dec 4, 2013

### Coco12

Is the answer that I got correct? Thanks for taking the time to answer. I know how to do it just wondering if it's correct

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4. Dec 4, 2013

### HallsofIvy

Staff Emeritus
Yes, y= x is the "liner oblique asymptote".

5. Dec 4, 2013

### Staff: Mentor

No, this isn't correct. It's the equation of the rational function you started with.

To find the oblique asymptote, either do what Mentallic suggested or carry out the long division to get x + a proper rational function. In a proper rational function, the degree of the numerator is less than that of the denominator.

6. Dec 6, 2013

Ok thank you