# Math Help - Backwards limit problem.

1. ## Backwards limit problem.

If $\lim_{x\to1}\frac{f(x)-8}{x-1}=10$ then find $\lim_{x\to1}f(x)$

This has me pretty stumped. Since the denominator is zero at x=1, I can't use limit of quotient is the quotient of the limits. My only other thought was to figure out f(x), which should be a polynomial where f(x)-8 has a factor of 1-x, but I can't figure out what that would be.

2. ## Re: Backwards limit problem.

Since the original limit is finite, what must

$\lim_{x\to 1}(f(x)-8)$ be?

3. ## Re: Backwards limit problem.

To do this problem, we're going to find out what f(x) is, and then take the limit. To find f(x), we do this:

We know that $f(x)-8$ has to have an $(x-1)$ in it. We also know that the other factor must be equal to 10, when we plug in a 1 for x. Therefore, the other factor must be $(x+9)$

If we FOIL, we have:

$(x-1)(x+9) = x^2+8x-9$

Now we separate the -8 from the mix:

$x^2+8x-1-8$

So we have:

$f(x) = x^2+8x-1$

When we plug in 1, we get our answer: 8

I hope this helps!

-Nathan

4. ## Re: Backwards limit problem.

Originally Posted by TutorMeNate
To do this problem, we're going to find out what f(x) is, and then take the limit.
I'm not sure I agree with this approach. f(x) doesn't even have to be defined in order to solve this problem.

5. ## Re: Backwards limit problem.

Originally Posted by Ackbeet
Since the original limit is finite, what must

$\lim_{x\to 1}(f(x)-8)$ be?
I'm guessing 8... So if we evaluated $\frac{f(x)-8}{x-1}$ by just plugging in x=1, should we get a fraction of $\frac{0}{0}$ because there is a finite limit?

6. ## Re: Backwards limit problem.

I'm not sure I agree with this approach. f(x) doesn't even have to be defined in order to solve this problem.
I figured there was another way to solve this problem, but to be honest, I did not know the exact method, so I used a way that I knew would work.

Sometimes the front door is locked, so you need to break through the window

-Nathan

7. ## Re: Backwards limit problem.

Originally Posted by beebe
I'm guessing 8... So if we evaluated $\frac{f(x)-8}{x-1}$ by just plugging in x=1, should we get a fraction of $\frac{0}{0}$ because there is a finite limit?
Looks good to me!

8. ## Re: Backwards limit problem.

Originally Posted by TutorMeNate
To do this problem, we're going to find out what f(x) is, and then take the limit. To find f(x), we do this:

We know that $f(x)-8$ has to have an $(x-1)$ in it.
No, we don't know that because we don't know that f is a polynomial.

We also know that the other factor must be equal to 10, when we plug in a 1 for x. Therefore, the other factor must be $(x+9)$

If we FOIL, we have:

$(x-1)(x+9) = x^2+8x-9$

Now we separate the -8 from the mix:

$x^2+8x-1-8$

So we have:

$f(x) = x^2+8x-1$

When we plug in 1, we get our answer: 8

I hope this helps!

-Nathan

9. ## Re: Backwards limit problem.

Good except for the "I'm guessing" part! No need to guess!

Thanks!

11. ## Re: Backwards limit problem.

Originally Posted by TutorMeNate
To do this problem, we're going to find out what f(x) is, and then take the limit. To find f(x), we do this:

We know that $f(x)-8$ has to have an $(x-1)$ in it. We also know that the other factor must be equal to 10, when we plug in a 1 for x. Therefore, the other factor must be $(x+9)$

If we FOIL, we have:

$(x-1)(x+9) = x^2+8x-9$

Now we separate the -8 from the mix:

$x^2+8x-1-8$

So we have:

$f(x) = x^2+8x-1$

When we plug in 1, we get our answer: 8

I hope this helps!

-Nathan
Wrong! Where does it say f(x) is a polynomial. We do not need to assume a polynomial to do this

Sorry did not see Hall's posts. Note to self: No more shooting from the hip (it don't harf hurt)

CB

12. ## Re: Backwards limit problem.

Oops, I apologize, I read the problem and started thinking polynomial:

My only other thought was to figure out f(x), which should be a polynomial where f(x)-8 has a factor of 1-x, but I can't figure out what that would be.
I should have thought it through more. Thanks for the correction!

-Nathan