1. Originally Posted by colby2152
...These definitions are meant for an Analysis course rather than an introductory Calculus course.
i think you meant that the other way around

other than that, the post is good. you neglected to mention how the signs affect the limit, but the examples you chose shows how to deal with them, so good job

looking forward to the next segment

i would also mention that neglecting the lower powers of x can work in a variety of environments, for example, under square roots and such.

for instance:

$\lim_{x \to \infty} \frac {3x}{\sqrt{4x^2 + 1}}$

we could treat this as $\lim_{x \to \infty} \frac {3x}{\sqrt{4x^2}} = \lim_{x \to \infty} \frac {3x}{2|x|} = \lim_{x \to \infty} \frac {3x}{2x} = \frac 32$

but be careful with this. sometimes dropping lower powers of x can come back to bite you. i saw a question like this the other day, i don't remember it now

2. Originally Posted by colby2152
I am not sure why, because when I enter the url address in my web browser, it comes up just fine. Could you try doing that for me to see if it works? (just so I can figure out this problem?)
That's probably because of the hotlink protection on the server. It protects the files to be accessed from other sites, but copying and pasting url to the bar works. If you have access to the file .htaccess, you can remove the protection. Or upload them to another server.

Originally Posted by Jhevon
indeed. copying and pasting the url in your message takes us to various pdf files. but when we click on the link itself, it sends us to freehostia.com

did you try clicking on it from your post?
Clicking on them can work for him because when you see the file once the browser caches it and when you try to reach the same file it doesn't download again. Clear your cache and then try to click on the links..

3. Originally Posted by wingless
That's probably because of the hotlink protection on the server. It protects the files to be accessed from other sites, but copying and pasting url to the bar works. If you have access to the file .htaccess, you can remove the protection. Or upload them to another server.
I found the problem wingless! Since my subscription on the host site is free, hot linking is not allowed. I wonder if a direct link to the website domain will work?

The Calculus Guru

4. Originally Posted by colby2152
I found the problem wingless! Since my subscription on the host site is free, hot linking is not allowed. I wonder if a direct link to the website domain will work?

The Calculus Guru
It works. Hotlink protection is usually for files, not pages.

5. Originally Posted by wingless
It works. Hotlink protection is usually for files, not pages.
In that case, I may just link it over to my lessons page which I will do this week!

6. ## Lesson 2.5 - Limits: Continuity

Lesson 2.5 - Limits: Continuity
Continuity is an important aspect of Calculus. We can show that a function is continuous in a variety of different ways. If a function is undefined at some point, then that point is a point of discontinuity.

Definition: A function $f(x)$ is continuous at $x=a$ if

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

A function is said to be continuous on an interval if it is continuous at every point within that interval.

Example 1.6-A

$f(x)=\frac{\pi}{x+2}$

Is this function continuous at $x = -2$?

$\lim_{x \rightarrow -2} \frac{\pi}{x+2} = DNE$

NO, this function is undefined at $x = -2$, and a vertical asymptote exists at such a point.

The definition of continuity can be turned around.

Corollary: If the function $f(x)$ is continuous at $x=a$, then:

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

The Pencil Rule
This is no official rule or theorem, but simply a rule of thumb to help you. Try graphing a function on a piece of paper. If you must lift your pencil for any moment off of the paper while graphing the function, then it certainly is not continuous. Basically, if you cannot trace the graph in one fell swoop, then it has a point of discontinuity somewhere.

Intermediate Value Theorem
If $f(x)$ is continuous on a closed interval $[a,b]$, and $c$ is any number between $f(a)$ and $f(b)$, then there is at least one number $x$ in the closed interval such that $f(x)=c$.

7. ## Lesson 2.6 - Limits: Squeeze Theorem

Lesson 2.6 - Limits: Squeeze Theorem
The squeeze theorem, also known as the pinching or sandwich theorem, is a theorem dealing with the limit of a function. It is used to compare one function to two other functions. If two functions approach the same limit at a point, and if the third function always has a value between these functions, then it has the same limit. This theorem is a highly useful method of comparison to identify limits.

Let $g(x) \le f(x) \le h(x)$ for some interval that all of these functions are defined on.

If a limit of $g(x)$ and $h(x)$ equal $L$, then that same limit of $f(x)$ equals $L$.

The functions $g(x)$ and $h(x)$ are said to be lower and upper bounds, respectively, of $f(x)$.

Example 2.6-A

$f(x)=x^2 sin\left(\frac{1}{x}\right)$

Find the limit of $f(x)$ as $x$ approaches zero. Using a numerical “plug and chug” approach gives us an undefined value. If we cannot graph it, then we must solve this limit algebraically, namely by using the squeeze theorem!

$g(x)=x^2$

$h(x)=(-x)^2$

$(-x)^2 \le x^2 sin\left(\frac{1}{x}\right) \le x^2$

$\lim_{x \rightarrow 0} (-x)^2=0$

$\lim_{x \rightarrow 0} x^2 = 0$

Therefore: $\lim_{x \rightarrow 0} x^2 sin\left(\frac{1}{x}\right) = 0$

8. ## Lesson 3.1 – Derivatives: Definition

Lesson 3.1 – Derivatives: Definition
The derivative is simply the rate of change within a function. That is what a derivative is in mathematics. After learning a few of the basic rules, it does not get much simpler than that.

As I said, the derivative measures the rate of change. In linear algebra, we call this the slope of a line. When measuring the derivative at a point, we can think of the concept of slope.

Example 3.1-A
Let us go through a rather long example to show the numerical analysis it takes to find a derivative if we were using only Algebra.

$f(x)=3x^3+4x$

Find the rate of change (derivative) at $x = 2$

Let’s find the slope for values close to $2$; between $x = 0$ and $x = 4$

$m=\frac{(f(4)-f(0))}{(4-0)}$

$m=\frac{(3(64)+4(4)-3(0)-4(0))}{4}$

$m=\frac{(192+16)}{4}$

$m=\frac{208}{4}$

$m=52$

Let’s find the slope for values even closer to $2$; between $x = 1$ and $x = 3$

$m=\frac{f(3)-f(1)}{(3-1)}$

$m=\frac{3(27)+4(3)-3(1)-4(1)}{2}$

$m=\frac{81+12-3-4}{2}$

$m=\frac{86}{2}$

$m=43$

One more time, and we will avoid using a calculator here; between $x = 1.5$ and $x = 2.5$

$m={f(2.5)-f(1.5)}{2.5-1.5}$

$m=3(2.5*2.5*2.5)+4(2.5)-3(1.5*1.5*1.5)-4(1.5)$

$m=3(6.25*2.5)+10-3(2.25*1.5)-6$

$m=3(6.25*(2+0.5) )+10-3(2.25*(1+0.5) )-6$

$m=3(12.5+3.125)+10-3(2.25+1.125)-6$

$m=3(15.625)+10-3(3.375)-6$

$m=46.875+10-10.125-6$

$m=40.75$

Our conclusion is that the slope, or value of the derivative, at $x = 2$ is approximately $40.75$. The true value is actually $40$. How do we find the true value of a derivative without doing many calculations?

The answer should be apparent at this point from what was already learned. To find what value the slope is approaching, we must use a limit. Let’s use the concept of slope to find this limit. Slope is computed as the “rise” over the “run”; y’s over x’s; output over input; or as the following in this limit:

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

$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

The more commonly known definition of the derivative is:
$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

9. Originally Posted by colby2152
$lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$
There is something interesting about this limit. If $f$ is defined in a neighborhood of $x$ and that limit exists, then $f$ is differenciable at $x$ and its limit is $f'(x)$. However, if $f$ is differenciable at $x$ we cannot conclude that limit exists. That is the interesting point, it does not work the other way around.

10. Originally Posted by ThePerfectHacker
There is something interesting about this limit. If $f$ is defined in a neighborhood of $x$ and that limit exists, then $f$ is differenciable at $x$ and its limit is $f'(x)$. However, if $f$ is differenciable at $x$ we cannot conclude that limit exists. That is the interesting point, it does not work the other way around.
That is intriguing. How did you conclude this? Remember I brought this up with you about a month ago? I wanted to show this definition as a modification of the actual definition to better fit the idea of slope and get across my point that slope = change = derivative.

11. ## Lesson 3.2 - Basic Properties & The Power Rule

Lesson 3.2 - Basic Properties & The Power Rule

Before we can discuss the various rules of differentiation, we need to introduce the symbolism and nomenclature of Calculus. There are a variety of ways to show the symbol of a derivative, but two notations are used the most.

The "prime" notation
This is the easiest way to show a derivative and is often the least complicated for students to grasp. Say we have a function, $f(x)$. It's derivative of the function with respect to $x$ is simply $f'(x)$. Say $y=f(x)$, then the derivative of $y$ with respect to $x$ is just [tex]y'. You can thank the simplicity of this notation to Joseph Louis Lagrange.

The "fraction" notation
This is the original notation used by Gottfried Leibniz, the cofounder of the mathematics known as Calculus. The derivative of the function $y$ with respect to $x$ can be written as: $\frac{dy}{dx}$ Leibniz's notation useful for differential equations because the symbol can be treated as a fraction is most cases.

In these lessons, both notations will be used.

BASIC PROPERTIES
1) Sum or Difference
$(f(x) + g(x))' = f'(x) + g'(x)$

$\frac{d}{dx}(f(x)-g(x)) = \frac{df}{dx} - \frac{dg}{dx}$

2) Coefficients
$(Cf(x))' = cf'(x)$
The coefficient can be factored out of the derivative.

These properties are the same as those of limits. However, the basic properties of limits that apply to multiplication, division and grouping of functions cannot be applied to derivatives.

At this point, you know one way of finding a derivative. It is long and tedious, so there must be some shortcuts - right?
There are about a half dozen shortcuts or rules that you will use frequently when finding a derivative. The first one to be learned is the power rule.

Whenever you have a polynomial of some form $x^n$, the derivative is simply $nx^{n-1}$. You simply bump that power to the side as a coefficient multiplier of $x$, and then drop the power of $x$ by one. Note that the base must be a variable. In this example, it is $x$. The only exception is when you have a constant. The derivative of a constant is ALWAYS ZERO!

Let's go through some examples...

Example 3.2-A
$f(x) = 7x^{100} + 5x^3 - 14$

We can go through this term by term thanks to our basic properties learned earlier in the section.

$f'(x)=100*7x^{100-1} + 3*5x^{3-1} - 0$
$f'(x)=700x^{99} + 15x^2$

Example 3.2-B
$h(t) = 4t^{3.7} - 7x^{-1}$

Do not let the negative or non-integer numbers confuse you. Continue with what we know...

$h'(t) = 3.7*4t^{3.7-1} - -(1)*7x^{-1-1}$
$h'(t) = 14.8t^{2.7} + 7x^{-2}$

Example 3.3-C
$f(\text{carrot}) = 4\sqrt{\text{carrot}}$

Change the radical to a fractional exponent for easy calculation...

$f(\text{carrot}) = 4\text{carrot}^{0.5}$

$f'(\text{carrot}) = 0.5*4 \text{carrot}^{0.5-1}$
$f'(\text{carrot}) = 2 \text{carrot}^{-0.5}$

Example 3.4-C
$d(k)= \frac{2k^5 - k^3 + 9}{k^2}$

We have not learned the product or quotient rules just yet, but we do not need to for such a problem. Use algebra to simplify the terms of the function.

$d(k) = \frac{2k^5}{k^2} - \frac{k^3}{k^2} + \frac{9}{k^2}$
$d(k) = 2k^3 - k + 9k^{-2}$

$d'(k) = 6k^2 - 1 - 18k^{-3}$

12. You can use $\lim$ (\lim) instead of $lim$ (lim).

Writing $carrot$ (carrot) as $\text{carrot}$ (\text{carrot}) looks better too.

13. ## Lesson 3.3 - Derivatives of Trig Functions

Lesson 3.3 - Derivatives of Trig Functions
For future reference, the easiest method to deal with trig is to break everything down into sines and cosines. It is simply less to committ to memory. The same can be said when it comes to finding derivatives. Nevertheless, knowing the derivatives of all six basic trigonometric functions can be helpful.

$sin(x)'=cos(x)$
$cos(x)'=-sin(x)$
$tan(x)'=sec^2(x)$
$cot(x)'=-csc^2(x)$
$sec(x)'=-sec(x)tan(x)$
$csc(x)'-csc(x)cot(x)$

As you can see, the derivatives of the the latter three trig functions are more cumbersome than the first three. Upon learning the chain rule, there will be no need to learn the derivatives of any of the trig functions except for sine and cosine. Also, derivatives of the hyperbolic trigonometric functions follow the same rules that their regular counterparts do.

The derivatives are periodic and have the same pattern every four derivatives as is shown with sine.

$f=sin(x)$
$f'=cos(x)$
$f''=-sin(x)$
$f^{(3)}=-cos(x)$
$f^{(4)}=sin(x)$

You can use this pattern as a method to find very high derivatives like the following.

Example 3.3-A
$h(t)=8cos(t)$

$h^{(37)}(t)=???$

You know that sine and cosine repeat every four derivatives, so any derivative can be divided by four with the remainder being the # of derivatives that are taken. (This is also known as taking modulo 4 of the number.)

$\frac{37}{4} = \frac{36}{4} + \frac{1}{4} \Rightarrow 9 + \frac{1}{4} \Rightarrow* 1$

Now, solve for the first derivative...

$h^{(37)}(t)=h'(t) \Rightarrow -8sin(t)$

14. Originally Posted by colby2152
The derivatives are periodic and have the same pattern every four derivatives as is shown with sine.
$f=sin(x)$
$f'=cos(x)$
$f''=-sin(x)$
$f^{(3)}=-cos(x)$
$f^{(4)}=sin(x)$
You may want to add that a simple method to find for example the 125th derivative of sin x would be to divide 125 by 4 repeatedly until you get a remainder smaller or equal to 4.

The remainder, lets call it $n$, then gives you the $n$-th derivative of sin.

So if we have 125th derivative... Divide by 4 , 31 times, until we have a remainder of 1.

So then it would be the first derivative of sin which is cos.

15. Originally Posted by janvdl
You may want to add that a simple method to find for example the 125th derivative of sin x would be to divide 125 by 4 repeatedly until you get a remainder smaller or equal to 4.

The remainder, lets call it $n$, then gives you the $n$-th derivative of sin.

So if we have 125th derivative... Divide by 4 , 31 times, until we have a remainder of 1.

So then it would be the first derivative of sin which is cos.
Good call, I added an example although I may add one more that has a coefficient in the parameter such as $sin(4w)$.

Page 4 of 7 First 1234567 Last