# not sure about what this limit represents

• May 14th 2008, 11:42 AM
i_zz_y_ill
not sure about what this limit represents
given f(x)=x^4 (iv) stuck
i) complete this identity f(x+h) = x^4+4x^3h+...
ii)simplif (f(x+h)-f(x))/h
iii)find lim h tends 0 of ii)
iv)state what this limit represents

the answer is gradient of (tangent to) curve but i'm not sure why. If someone could explain to me the reasoning then i could apply it to other q's,thnx
• May 14th 2008, 11:46 AM
Mathstud28
Quote:

Originally Posted by i_zz_y_ill
given f(x)=x^4 (iv) stuck
i) complete this identity f(x+h) = x^4+4x^3h+...
ii)simplif (f(x+h)-f(x))/h
iii)find lim h tends 0 of ii)
iv)state what this limit represents

the answer is gradient of (tangent to) curve but i'm not sure why. If someone could explain to me the reasoning then i could apply it to other q's,thnx

Because $f'(x)=\lim_{\Delta{x}\to{0}}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}$

Imagine a secant line intersecting the curve you are looking for in two places.

Now let the distance between those places be denoted $\Delta{x}$..

Next imagine drawing the line differently so you decrease $\Delta{x}$...do this again and again and again

Eventually as you draw the line so that $\Delta{x}\to{0}$ the line intersects the curve at ONE point...and at that point the line shares the slope of the curve...

Thus this is the derivative of a function(e.g. the slope)
• May 14th 2008, 11:54 AM
Moo
Hello,

Quote:

Originally Posted by i_zz_y_ill
given f(x)=x^4 (iv) stuck
i) complete this identity f(x+h) = x^4+4x^3h+...
ii)simplif (f(x+h)-f(x))/h
iii)find lim h tends 0 of ii)
iv)state what this limit represents

the answer is gradient of (tangent to) curve but i'm not sure why. If someone could explain to me the reasoning then i could apply it to other q's,thnx

That's right (a little vague though)! (Wink)

Now, why ?
[non formal explanation]
Because the gradient corresponds to the progression of the ordinates with respect to the absciss, when dealing with a curve.
The gradient also represents the "speed" of evolution for the ordinates with respect to the absciss.

Now, this is clearly the quotient of (difference of ordinates) by (difference of absciss), which corresponds to a gradient.

But what does it mean when taking the limit of h to 0 ? This means that we're dealing with a very very small interval of the curve, and calculating the "instantaneous" or "local" gradient.

You can try, if you've already dealt with some physics, to get an equivalence, or just try to represent it to yourself, with average speed (gradient) and instantaneous speed :
- if you drive a car, you go from A to B in a time from t1 to t2. The average speed will be the distance between A and B (which is equivalent to a difference of coordinates) divided by the time it took : t2-t1. This is a gradient.

- now, I'm sure you were tired at moments, so you drove slower. The speed was not always equal to the average speed. That's here that the "instantaneous" speed intervenes : if you take the speed over a very little time, it's like finding the gradient over a very small interval.

If you want to know more about the exact term, search for "derivative" on google (Wink)

Dunno if it's clear or not... I'm more used to explaining it in French :'(
• May 14th 2008, 12:52 PM
i_zz_y_ill
uhuh informal, formal gotcha!
k well basically i have no idea what your both on about so i just bunged in x^4 and (x+4)^4 - f(x) divided by 4 in my calculatorand this graph looked like the graph of its gradient if u know what i mean!!! gradient decreasing from -1 to 0 etc,,,
I guess youre both talking about that in some sense, thnx for reminder it was clear and my english is probably worse than yours!
• May 14th 2008, 01:28 PM
Plato
$\frac{{f(x + h) - f(x)}}{h} = \frac{{\left( {x^4 + 4x^3 h + 6x^2 h^2 + 4xh^3 + h^4 } \right) - x^4 }}{h} = 4x^3 + 6x^2 h + 4xh^2 + h^3
$