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

In summary, the difference between \Delta V and dV can be represented by \varepsilon \Delta x, where \varepsilon \to 0 as \Delta x \to 0. This can be seen by simplifying the equations and observing that \varepsilon is a function of \Delta x, which approaches 0.
  • #1
GunnaSix
35
0

Homework Statement



Let [tex]V=x^3[/tex]

Find [tex]dV[/tex] and [tex]\Delta V[/tex].

Show that for very small values of [tex]x[/tex] , the difference

[tex]\Delta V - dV[/tex]

is very small in the sense that there exists [tex]\varepsilon[/tex] such that

[tex]\Delta V - dV = \varepsilon \Delta x[/tex],

where [tex]\varepsilon \to 0[/tex] as [tex]\Delta x \to 0[/tex].

Homework Equations



[tex]dV = 3x^2 dx[/tex]

[tex]\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3[/tex]

The Attempt at a Solution



I worked it down to

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

Can I say that [tex]\lim_{\Delta x \to 0} \Delta x = dx[/tex] so that

[tex]\lim_{\Delta x \to 0} \varepsilon = 3x(0) + (0)^2 + 3x^2(1-1) = 0[/tex] ?
 
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  • #2
GunnaSix said:

Homework Statement



Let [tex]V=x^3[/tex]

Find [tex]dV[/tex] and [tex]\Delta V[/tex].

Show that for very small values of [tex]x[/tex] , the difference

[tex]\Delta V - dV[/tex]

is very small in the sense that there exists [tex]\varepsilon[/tex] such that

[tex]\Delta V - dV = \varepsilon \Delta x[/tex],

where [tex]\varepsilon \to 0[/tex] as [tex]\Delta x \to 0[/tex].


Homework Equations



[tex]dV = 3x^2 dx[/tex]

[tex]\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3[/tex]


The Attempt at a Solution



I worked it down to

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

Can I say that [tex]\lim_{\Delta x \to 0} \Delta x = dx[/tex] so that

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

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

[tex]
dV = 3x^2 dx [= 3x^2\Delta x]

[/tex]
[tex]
\Delta V = 3x^2 \Delta x + 3x (\Delta x)^2 + (\Delta x)^3
[/tex]

you have:

[tex]
\Delta V-dV= 3x (\Delta x)^2 + (\Delta x)^3 = (\Delta x)(3x\Delta x + (\Delta x)^2))= \Delta x (\varepsilon)
[/tex]

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

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

1. What is the definition of a limit?

A limit is a mathematical concept that describes the behavior of a function as its input approaches a certain value or point.

2. How is the limit of a function evaluated?

The limit of a function can be evaluated by plugging in values that approach the given point or by using algebraic techniques such as factoring and simplifying.

3. Why is the limit of a function important?

The limit of a function helps us understand the behavior of the function and predict its output at a given point. It also plays a crucial role in the development of calculus and other mathematical concepts.

4. Can the limit of a function exist at a discontinuity?

No, the limit of a function cannot exist at a discontinuity. This is because the function is undefined at that point and cannot be evaluated.

5. How is the limit of a function different from the value of the function at a given point?

The limit of a function describes the behavior of the function as it approaches a certain point, while the value of the function at a given point is just a single output value. The limit takes into account the behavior of the function as a whole, while the value only considers a specific point.

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