Can someone please show me the steps to complete a limit function?

Question

lim f(x(sub 0)+h)-f(x(sub 0))/h
h-->0

f(x)=3x^2
x(sub 0)=1

Thanks.

Thanks, that helped a lot...but where did you get the three from when you plugged in the values before simplifying? Did it just come out of the f(x)=3x^2

Answer 1

First, you plug in your variables:
[h→0]lim (3(1+h)² - 3(1)²)/h
Now simplify:
[h→0]lim (3(1+2h+h²) - 3)/h
[h→0]lim (6h+3h²)/h
And cancel:
[h→0]lim 6+3h

Of course, as h→0, that second term also approaches zero and therefore vanishes, so we may now evalutate this limit as:

6

Edit: "Thanks, that helped a lot...but where did you get the three from when you plugged in the values before simplifying? Did it just come out of the f(x)=3x^2"

Yes, that's exactly where it came from. When you evaluate an expression such as f(x_0+h), you basically put the function f there, substituting the stuff inside the parentheses everywhere you see the x (or whichever variable is in the parentheses when the function is defined - for instance if you were given f(a)=3a², you would evaluate f(x_0+h) by putting (x_0+h) everywhere you see the a - this would also give 3(x_0+h)²).

Answer 2

Are you sure you got that right? you probably want:

Lim [f(x0+h) - f(x0)] / h as h->0

The way you wrote it, the limit is [Negative Infinity]
the way I wrote it, the limit is:

f(x+h) = f(x) + df/dx h = 3x^2 + 6x h (for small h)
so [f(x+h) - f(x)] / h = [6 x h] / h = 6 x and at x = 1 you get

Lim(h->0) = 6


-jose-

Answer 3

well, because you have an h in there, you have to solve it using the definition of limit: (f(x+h)-f(x))/h
plug in everything and try to solve.
also, its sort of like a composition of functions-whereever you see an x, plug in (x+h) for the first part, and just write it out again as f(x)
hope this isnt confusing..lol