# Thread: Evaluating Limits

1. ## Evaluating Limits

Hi, I'm new here and this is my first post but I'm having trouble with one of my homework problems. I have the function:

g(x) = (a(x+b)(x-c)(x+d)) / (e(x-f)(x-b)(x-c)), where a, b, c, d, e, and f are all unique, positive real numbers.

a) At what values of x=h does the limit g(x) = -∞ or +∞ as x->h?

b) Evaluate the limit of g(x) as x->c.

c) Does the limit g(x) as x->∞ exist?

I'm completely stuck and unsure where to start...

2. ## Re: Evaluating Limits

Originally Posted by daveMathews48
g(x) = (a(x+b)(x-c)(x+d)) / (e(x-f)(x-b)(x-c)), where a, b, c, d, e, and f are all unique, positive real numbers.
a) At what values of x=h does the limit g(x) = -∞ or +∞ as x->h?

b) Evaluate the limit of g(x) as x->c.

c) Does the limit g(x) as x->∞ exist?

I'm completely stuck and unsure where to start...
−∞−∞
This is very good question to teach you about limits.

The answer to a) is $f~\&~b$. Now you reply giving the reasons.
Then maybe we can do more.

3. ## Re: Evaluating Limits

If I cancel out the (x-c) from both the top and bottom then set the new denominator e(x-f)(x-b) to 0? Wouldn't that leave me with e, f, and b?

4. ## Re: Evaluating Limits

Originally Posted by daveMathews48
If I cancel out the (x-c) from both the top and bottom then set the new denominator e(x-f)(x-b) to 0? Wouldn't that leave me with e, f, and b?
No it does not. Because $e$ is a positive number.

5. ## Re: Evaluating Limits

Ah that's right, thanks. So for part b with (x-c) cancelled out, how do I evaluate that limit?

6. ## Re: Evaluating Limits

For part b) the answer is a fraction. The numerator of which is $a(c+b)(c+d)$.
What is its denominator?

7. ## Re: Evaluating Limits

Are you substituting c for x? That would make the denominator e(c-f)(c-b)?

8. ## Re: Evaluating Limits

Also, I was thinking about part c. Doesn't lim g(x) as x-> ∞ look for values where x is really large? If I expand the numerator and denominator I get:

(a(x^2+bx+dx+bd)) / (e(x^2-bx-fx+bf)), which for large x leaves me with ax^2/ex^2? Does that mean the limit exists and is a/e?

9. ## Re: Evaluating Limits

Originally Posted by daveMathews48
Are you substituting c for x? That would make the denominator e(c-f)(c-b)?
In limits NEVER JUST SUBSTITUTE!. Ask what happens near the limiting value?
If $x$ is near $c$ then $(x-b)$ is near $c-b$.

Now for the last part. We play the 'back-of-the-book" game.
You look up the answer. You see: "the limit exists and equals $\frac{a}{e}$".
You may be scratching your head why? Can you explain why?

10. ## Re: Evaluating Limits

Ah ok, that makes sense. Well, I don't know if that's the right answer this is one of those problems that's not in the back of the book :P. I mean I understand I can reduce the fraction that way because when x is really large, x^2 is much more important than anything else in the equation. And when I have x^2/x^2 that just equals 1.