Is the Difference Between \Delta V and dV Very Small for Small \Delta x?

[GunnaSix]

Homework Statement

Let V=x^3

Find dV and \Delta V.

Show that for very small values of x , the difference

\Delta V - dV

is very small in the sense that there exists \varepsilon such that

\Delta V - dV = \varepsilon \Delta x,

where \varepsilon \to 0 as \Delta x \to 0.

Homework Equations

dV = 3x^2 dx

\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3

The Attempt at a Solution

I worked it down to

\varepsilon = 3x \Delta x + (\Delta x)^2 + 3x^2 \left(1 - \frac{dx}{\Delta x} \right)

Can I say that \lim_{\Delta x \to 0} \Delta x = dx so that

\lim_{\Delta x \to 0} \varepsilon = 3x(0) + (0)^2 + 3x^2(1-1) = 0 ?

 

[LCKurtz]

Homework Statement

Let V=x^3

Find dV and \Delta V.

Show that for very small values of x , the difference

\Delta V - dV

is very small in the sense that there exists \varepsilon such that

\Delta V - dV = \varepsilon \Delta x,

where \varepsilon \to 0 as \Delta x \to 0.

Homework Equations

dV = 3x^2 dx

\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3

The Attempt at a Solution

I worked it down to

\varepsilon = 3x \Delta x + (\Delta x)^2 + 3x^2 \left(1 - \frac{dx}{\Delta x} \right)

Can I say that \lim_{\Delta x \to 0} \Delta x = dx so that

\lim_{\Delta x \to 0} \varepsilon = 3x(0) + (0)^2 + 3x^2(1-1) = 0 ?

You basically have it. You need to note that \Delta x and dx are the same thing, regardless of limits. Using your equations:

<br /> dV = 3x^2 dx [= 3x^2\Delta x]<br /> <br />
<br /> \Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3<br />

you have:

<br /> \Delta V-dV= 3x (\Delta x)^2 + (\Delta x)^3 = (\Delta x)(3x\Delta x + (\Delta x)^2))= \Delta x (\varepsilon)<br />

Does this \varepsilon work? Note that you don't need to take any limits to answer the question. You just need to observe that \varepsilon \rightarrow 0 as \Delta x does.