# limit and derivatives - precise definitions

Can someone explain the following definitions of a limit and derivatives. Issac did a good job explaining it in the previous thread: http://math.stackexchange.com/questions/8688/infinite-series-archimedean-principle

The definition of a limit is defined as the following:

$lim_{x \to a} f(x) = T$, what we actually mean is if we take any number M > T, then we can force $f(x) < M$ by taking $x$ sufficiently close to $a$. Similarly, if we take any $L < T$, then we can force $f(x) > L$ by taking $x$ sufficiently close to $a$.

The definition of a derivative is defined as the following: The derivative of $f$ at $a$ is the value denoted by $f'(a)$, such that for any $L < f'(a)$ and any $M > f'(a)$, we can force $L < \frac{f(x)-f(a)}{x - a} < M$ simply by taking $x$ sufficiently close to (but not equal to $a$).

This is my first real analysis course and I am very confused. If you choose to use episilon delta, please try to make it as simple as possible.

thanks

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It's pretty straightforward, I think, or, as they say in TV Tropes, "exactly what it says on the tin". Think of it as a game: there's a limit/derivative if and only if you can such values of $L$ and $M$. You should try it yourself with simple functions and specific values of $L$ and $M$. – J. M. Nov 4 '10 at 22:11
bad question. This is the wrong way to think about analysis. – anon Nov 5 '10 at 17:09
Its not me, its the bad book we've been given. Can you recommend a good book? – Tyler Hilton Nov 5 '10 at 18:55
This was enlightening to me, maybe you could take a look at it. – J. M. Nov 5 '10 at 22:51
That is the same book we are using in class ! – Tyler Hilton Nov 6 '10 at 18:19

Here's how I would explain limits:

Think of the function $f(x) = x^{2}$. We might want to ask ourselves: "What is $\lim_{x \rightarrow 3} f(x)$?" Put simply: As $x$ gets near 3, where happens to our function? Clearly, our function goes to 9, which is $3^2$.

One must understand, however, that a limit is a much more subtle argument. In our case, we could just "plug in" 3. But suppose we wanted to take $\lim_{x \rightarrow 0} \frac{1}{x}$ or $\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$. In these cases, we cannot simply "plug in" our value for x, and we must understand what a limit really is.

The arrow $\rightarrow$ in the limit sign means "approaching." In our first example, we wanted to find out what happens as x approaches 3. Let's look at this ourself:

Suppose $x=1$, then $x^2=1$.

Suppose $x=2$, then $x^2=4$.

Suppose $x=2.5$, then $x^2=6.25$.

Suppose $x=2.9$, then $x^2=8.41$.

Suppose $x=2.99$, then $x^2=8.9401$.

As x gets closer and closer to 3, $x^2$ (our function) gets closer and closer to 9. We can also come from the other direction.

Suppose $x=5$, then $x^2=25$.

Suppose $x=4$, then $x^2=16$.

Suppose $x=3.5$, then $x^2=12.25$.

Suppose $x=3.1$, then $x^2=9.61$.

Suppose $x=3.01$, then $x^2=9.0601$.

Now, let me introduce the terminology of epsilons $\epsilon$ and deltas $\delta$. The delta is how far away we are from our chosen limit value for x. In this case, the delta was how far away we were from 3. When we set x=3.5, delta was equal to 0.5. When x=3.1, delta was equal to 0.1.

Correspondingly, epsilon was how far away we were from the value of the limit: 9. When we set x=3.1, and our delta was 0.1, we got $f(x) = 9.61$, so the epsilon was 0.61.

And therein lies the fundamental idea of a limit. We can make the epsilon smaller simply by shrinking the delta. The closer your x is to 3, the closer your f(x) is to 9. If you wanted to be within 0.001 of 9 (i.e. $\epsilon = 0.001$), you'll need to be really close to x, and have a really small $\delta$. To be precise, $\delta = 0.00017$.

We therefore say that the limit of $f(x) = x^2$ as x approaches 3 is equal to 9, because we can make the output of the function be as close to 9 as we want, simply by constricting the input to be close to 3.

The above picture provides a great visual description. We have some curvy function $f(x)$, and we want to say that the limit at $x=a$ is equal to $L$. This can be shown because if we narrow the lines $a-\delta$ and $a+\delta$, thereby narrowing our input, we narrow the lines $L+\epsilon$ and $L-\epsilon$, constricting our output. We therefore must conclude that the limit is $L$, because the value of our function approaches $L$ and gets closer and closer to it the closer we get to $x=a$.

I hope this explanation has helped.

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hey you are wonderful! This is not meant to be rude against the other answerers. I just have a question. My prof talks about "no matter how small episilon we choose, we can always find a delta". its midnight here, and im gonna re -read your comment tomorrow morning. – Tyler Hilton Nov 5 '10 at 4:08
i think in your last line of "suppose" its supposed to be x = 3.01? – Tyler Hilton Nov 5 '10 at 4:19
Yeah. Fixed, thanks. – Isaac Solomon Nov 5 '10 at 4:55
My prof talks about "no matter how small epsilon we choose, we can always find a delta". What this means (in the context of our above example), is that now matter how close I want to get to 9, we can always get there by letting $x$ get close enough to 3. Generally speaking, in the context of the above picture, no matter how narrow I want to make my $L+\epsilon$ and $L-\epsilon$ (by shrinking my $\epsilon$), I can get f(x) to lie within them by shrinking $\delta$, so that $a+\delta$ and $a-\delta$ get closer together, thereby limiting my possible input values for x to be very close to 3. – Isaac Solomon Nov 5 '10 at 5:01

I think it's easier if you think about it visually.

With loss of generality, we can assume as we get closer to 'a' our function is increasing, it could be constant, decreasing, or whatever.

Let's say you have X1 < a < X2 on the X axis with the corresponding f(X1)=L < T < M=f(X2)
[Notice that f(a) doesn't necessarily equal T]

If we start squeezing X1 and X2 closer together and hence closer to a, we are forcing f(X1) and f(X2) closer to T. Therefore for x < X2, f(x) < f(X2)=M.
Similarly, for x > X1, f(x) > f(X1)=L

I hoped this helped. When in doubt, draw a picture.

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