# Math Help - Confused between deriviative of a Limit and a limit

1. ## Confused between deriviative of a Limit and a limit

The formal definition of a limit is that as x approaches zero, the value of x is f(x).

Then what is the deriviative of a limit? (I know the equation, but what does it mean)

I guess what I'm trying to say is....what is a limit?

The book we are using is absolutely useless.

2. Also the derivative of a limit . . . a limit normally ends up being a constant number, the derivative of a constant . . . is 0.

Um . . . the formal definition for a limit that pertains to functions is not exactly that.

Limit Def:

Let f(x) be a function defined on an interval that contains x=a, except possibly at x=a. Then we say that,

if and only if for all $\epsilon > 0$ there is a number $\delta > 0$ such that

$0<|x-a|<\delta \text{ implies } |f(x)-L|<\epsilon$

then we say:

$\lim_{x \rightarrow a}f(x)=L$

The derivative on the other hand is very different.

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

3. The formal definition of a limit is that as x approaches zero, the value of x is f(x). That is a false statement.

what is a limit? If the limit exists, then the limit is a number.

Then what is the deriviative of a limit? Since the limit is a number its derivative is zero.

4. Ok, I'm confused.

"Let f(x) be a function defined on an interval that contains x=a, except possibly at x=a. Then we say that,"

x=a, where did you get a? did you mean that a was a constant?

What is a limit exactly? (english definition, please, in simpler terms). If the professor were to ask me what is a limit, what would you say?

I know what a derivative is (the derivative of velocity is acceleration etc) but how does this relate to the Limit?

5. Consider this: $f(x) = \frac{{\sin (x)}}{x},\;\,\lim _{x \to 0} f(x) = 1\;\& \,f(0)\text{}$ is not defined.
That why your first statement is false.

The meaning of $\lim _{x \to a} f(x) = L$ can easily be explained in plain English.
For x ‘close to’ a, but not equal to a, then the value of the function at x is close to L.
In symbols, $\left( {\forall x} \right)\left[ {x \approx a\;,\,x \ne a\, \Rightarrow \,f(x) \approx L} \right]$.
In the above example $f(x) = \frac{{\sin (x)}}{x},\;\,\;\& \,f(0)$ is not defined.
However, whenever a is close to 0 then $f(a)$ is close to 1.

It’s our job in mathematics to make concepts such as ‘close to’ precise.

6. In the definition the a is just a number inside the interval you're looking at. So like if you're looking at 0 to 10, a could be any number inside there.

$\lim_{x \rightarrow a} f(x)=?$

The line above is basically asking where is the function going when it gets close to the number a.

So say like my function is:

$
f\left( x \right) = \left\{ \begin{gathered}
x, for -\infty 5, for \text{ } x = 2 \hfill \\
x, for \text{ } 2 \end{gathered} \right.
$

So then if i'm looking at this function.

$\lim_{x \rightarrow 2} f(x)=?$

It's asking where is the function heading to when it gets close to 2.

Well, you can use the formal definition, or you could use the not so formal way and just draw the graph. For the most part the graph looks like y=x, except when x = 2, f(x) = 5.

So when you look at it the function is heading to 2 right?

well you can also check that by plugging in numbers really close to 2. (the empirical method)

Like plugging in 1.9, 1.99, 1.9999, and then looking at it from the other end, 2.01, 2.001, 2.00001.

Though the empirical method doesn't always work it gives a pretty good indication of the limit of your function.

So your derivative, well the derivative is basically looking at the slope of your function, only it's concerned with what the slope of your function is at any given point, so it's the instantaneous slope.

The derivative is related to a limit in the sense that the derivative is defined by a limit.

So the slope of a function f(x)line:

$\frac{\Delta f(x)}{\Delta x}$

$\frac{f(x_2)-f(x_1)}{\Delta x}$

However since: $x_2-x_1=\Delta x$ which is just looking at the difference between 2 points.

You can always just set any x $x = x_1$

Then $x_2$ becomes $x+\Delta x$ because you just add the difference.

So then you get:

$\frac{f(x+\Delta x)}{\Delta x}$

Then the derivative is when:

$f'(x)=\lim_{ \Delta x \rightarrow 0}\frac{f(x+\Delta x)}{\Delta x}$

in other words, when I take the slope, I take 2 different points, the derivative says what happens when the difference between these 2 points shrink to 0? well you get the same point, so then you're looking at the slope of your function at THAT specific point. and that's what your derivative is, the instantaneous slope.

7. seld explained it like my book...hard to understand, the examples are too complex, lengthy etc.

Plato explained it perfectly

Thanks both of you.

8. I have another question.

Why would you take the derivative of a limit? Is it because you have a line tangent to a curve and you want to find the slope of that line at a point in the curve. So you use the derivative of a limit?

9. the derivative is the instantaneous slope, in other words it's the slope at any point on the line that you're looking at.

If you have a tangent line to a curve at some point say a parabola, the slope of the tangent line is the derivative at that point.

So if: y = x^2.

derivative of x^2 is 2x.

So if you take a tangent line to your parabola say at the point (1,1)

the slope of the tangent line is the derivative and the derivative is 2x the slope will be:

2(1)=2.

You can try it out.

I've no idea why you'd want to take the derivative of a limit, because as plato stated earlier if it exists, it's a number. Derivatives of numbers are 0. So if the limit exists the derivative would always be 0, which . . . seems like it's nonsensical to take the derivative of it.