# Limit of Delta x question

1. Jan 18, 2010

### GunnaSix

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

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 $$\Delta x \to 0$$.

2. Relevant equations

$$dV = 3x^2 dx$$

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

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

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

2. Jan 18, 2010

### LCKurtz

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

$$dV = 3x^2 dx [= 3x^2\Delta x]$$
$$\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3$$

you have:

$$\Delta V-dV= 3x (\Delta x)^2 + (\Delta x)^3 = (\Delta x)(3x\Delta x + (\Delta x)^2))= \Delta x (\varepsilon)$$

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.