# A Basic Function Problem with a Limit

1. Sep 24, 2009

### DMOC

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

For all real numbers x and y, let f be a function such that f(x+y)=f(x)+f(y)+2xy and such that the limit as h approaches zero is f(h)/h=7. $$\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7$$

a) Find f(0) and justify your answer.

b) Use the definition of the derivative to find f ' (x).

c) Find f(x).

NO CALCULATOR USE ALLOWED!

2. Relevant equations

Limit as h->0 is (f(x+h)-f(x))/h

3. The attempt at a solution

To solve part a, I looked at the original equation. To find f(0), I just put in 0 for x and 0 for y.

Let x=y=0.

And thus, I got f(0)=0, so now part a should be cleared. Right?

Then I get to part b and I'm a little stuck. Find f ' (x) ? Should I be using the equation I put in the relevant equations section to help me out with this?

Last edited: Sep 24, 2009
2. Sep 24, 2009

### aPhilosopher

The definition of the derivative is the equation that you gave, so yes.

3. Sep 24, 2009

### DMOC

OKay, so here goes my shot at part b (do you think part a is right?):

f ' (x) is ....

(f(x+7) - f(x)) / 7

Hm ... I'm getting a little lost here. I just substituted h for 7.

Also, is there a way to post these fractions using the latex reference?

4. Sep 24, 2009

### aPhilosopher

$$\stackrel{Lim}{h\rightarrow0}\frac{f(x + h) - f(x)}{h}$$

What can you say about f(x + h)?

to do the fractions, just use \frac{}{} where the numerator goes in the first bracket pair and the denominator goes in the second.

5. Sep 24, 2009

### DMOC

Thanks for letting me know about the fractions part.

For f(x+h), I was thinking that would just be equal to f(x) since h is approaching zero. However ...

$$\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7$$

Which doesn't make sense anyway since that would be undefined as h approaches zero. Hm ....

Last edited: Sep 24, 2009
6. Sep 24, 2009

### aPhilosopher

You have the formula for f(x + [strike]y[/strike]h) in your original post ;)

7. Sep 24, 2009

### aPhilosopher

Also, $$\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7$$ according to your original post. You edited it but left the old version. Which one is it?

8. Sep 24, 2009

### DMOC

Sorry, the limit as h goes to zero is definitely SEVEN.

So how about this. I was thinking that since I could do this ....

f(x+y) = f(x) + f(y) + 2xy
f(x+y) - f(x) = f(y) + 2xy

Then couldn't I say that since h is approaching zero, then it figures that:

x=y=0=h as well. I can then substitute the h's in for x or y?

Am I on the right track here?

9. Sep 24, 2009

### aPhilosopher

well f(x+h) - f(x) = f(h) + 2xh is right but that's inside a limit and over h as h goes to zero.

now, as I'm sure you know, the sum of limits is the limit of sums. So do it like that. There is no y in this equation and you can't set x to a value because you're taking the limit for all x so it needs to remain a variable.

10. Sep 24, 2009

### DMOC

I'm not sure what to do what you said, because we never discussed limits in my class yet.

11. Sep 24, 2009

### aPhilosopher

Ok. I don't think there's a way to do this without limits as they're in the problem as stated so the rule you want is:
$$\lim_{h\to a}(f(h) + g(h)) = \lim_{h\to a} f(h) + \lim_{h\to a} g(a)$$

See if that helps.

12. Sep 24, 2009

### DMOC

I think this problem requires knowledge of limits, although I never learned that last year (I think my teacher expected us to know from last year).

So for part b:

f ' (x) = $$\lim_{h\to 0} (f(x+h)/h) + \lim_{h\to 0} f(x)$$

Hopefully this is on the right track.

13. Sep 24, 2009

### aPhilosopher

That's the idea but do you remember how to expand f(x + h)?

Just keep it in one limit until you have expanded f(x + h) and the f(x) cancels out like it did before. And remember that when a sum is on top of a fraction, the denominator goes with both sides of the sum if you split it up!

$$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$$

14. Sep 24, 2009

### DMOC

Revised version.

f ' (x) = $$\lim_{h\to 0} ((f(x+h)/h) + f(x/h))$$

15. Sep 24, 2009

### DMOC

And, no I do not know how to expand f(x + h). I know this is easy once I understand the concept but I'm really new to it, sorry.

16. Sep 24, 2009

### aPhilosopher

That's ok. Look at your original post and post #8 and #9 in this thread. You've already done it

17. Sep 24, 2009

### DMOC

Really?

So do I substitute f(h) + 2xh for f(x+h) - f(x) now?

I'm still a bit confused, sorry.

18. Sep 24, 2009

### aPhilosopher

No problem.

Try it and see what happens. Post your results!

19. Sep 24, 2009

### DMOC

$$\stackrel{Lim}{h\rightarrow0}\frac{f(y) + 2xy}{h}$$

Do I put an equal sign after or before the limit part?

20. Sep 24, 2009

### aPhilosopher

Ok, where is the y coming from?