1. ## Limit question

Determine the value of a so that $f(x) = \frac{x^2+ ax + 5}{x+1}$has a slant asymptote y = x + 3

2. ## Re: Limit question

You want the following limit to exist and equal zero:

$$\lim_{x \to \infty} [f(x) - (x+3)] = 0$$

Solve for $a$. You want the numerator to be a constant.

3. ## Re: Limit question

$$f(x) = \frac{x^2+ ax + 5}{x+1} = (x+3) + \frac{b}{x+1} \, \text{(b constant)}$$

4. ## Re: Limit question

So a=4 and b=2

5. ## Re: Limit question

Yes, that is correct. For x not equal to 3, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}= x+ 3$.

6. ## Re: Limit question

Originally Posted by HallsofIvy
Yes, that is correct. For x not equal to 3, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}= x+ 3$.
Small typo:

For x not equal to 1, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}+\frac{2}{x+1}= x+ 3 + \frac{2}{x+1}$.

7. ## Re: Limit question

Do you mean "for X not equal to -1"?

8. ## Re: Limit question

Originally Posted by Archie
Do you mean "for X not equal to -1"?
Lol, good point. Fixed.

9. ## Re: Limit question

Now it says "X not equal to 1", not "X not equal to -1"

10. ## Re: Limit question

Originally Posted by SlipEternal
Small typo:

For x not equal to 1, $\displaystyle \frac{x^2+ 4x+ 5}{x+ 1}= \frac{(x+ 1)(x+ 3)}{x+ 1}+\frac{2}{x+1}= x+ 3 + \frac{2}{x+1}$.
Yes. Thank you and Archie.