1. ## Limit problem

if y= 2x^2 - 3

limit
h->0

f(x=h) - f(x)
------
h

should be on the same line but wont format correctly

2. hand i also have to explain the meaning of the answer

3. Instead of expecting us to show the entire solution and explain every step, you should be showing that you've made some effort to at least attempt this and have shown exactly where you are stuck. You should also have made some effort to post the question correctly and learn some LaTeX so that you can format it, instead of expecting us to decipher it.

To start, I expect you meant to find

$\displaystyle \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$

where $\displaystyle f(x) = y = 2x^2 - 3$.

Have you at least tried to evaluate $\displaystyle f(x + h)$, substitute these functions into the difference quotient and simplify?

4. okay i havent done limits on about... 8 months

ok i can evaluate that to get

lim
h-> 0

2x^2 + 4xh + h^2 - 3 - 2x^2 + 3
---
h

then to

4xh + h^2
---
h

but whats the answer? see doing it i can do but giving answer i cant

sorry for being rude.. =/

5. Not quite...

$\displaystyle f(x + h) = 2(x + h)^2 - 3$

$\displaystyle = 2(x^2 + 2xh + h^2) - 3$

$\displaystyle = 2x^2 + 4xh + 2h^2 - 3$

6. will the line

$\displaystyle (4xh + 2h^2)/h$

be better then?

7. Correct. You should now see that there is a common factor in the numerator. The next step should be obvious...

8. now see heres the problem. I don't actually know the point of limits. see here i get stuck. do i say what x is or what h is?

9. You haven't simplified enough. Start by taking a common factor, and simplifying...

Then, you will be able to find this limit - which is basically what function this becomes when you make $\displaystyle h$ go to $\displaystyle 0$...

10. so you mean that when simplified its
$\displaystyle 4x + 2h$

and when i take h as 0 the answer is 4x?

what does that mean though

11. Correct, well done.

What you have done is found a new function that will give you the GRADIENT of your original function at ANY point on that original function.

12. so your saying that i simple differentiated $\displaystyle y=2x^2 -3$

wow thats really confusing

13. Yes, that is what you have done.

This is the DEFINITION of the derivative. But since it's incredibly time consuming, mathematicians have found rules for differentiating functions.