Difference between revisions of "Calculus"
(Created page with "== Limits == Definition of a limit: <math>\lim_{x \rightarrow c} f(x) = L\,</math> means that for all real values of <math>\epsilon > 0 \,</math>, there exists <math>\delta...") |
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Now, if we want to prove that it is L, we have to look at ''all'' possible numbers, especially small ones. Now, let's choose a number x that is awfully close to c, and see what f(c) gives us. It should get use well below this number we imagined. | Now, if we want to prove that it is L, we have to look at ''all'' possible numbers, especially small ones. Now, let's choose a number x that is awfully close to c, and see what f(c) gives us. It should get use well below this number we imagined. | ||
+ | |||
+ | === Example === | ||
As a simple example, let's find: | As a simple example, let's find: | ||
− | <math>\lim_{x \rightarrow | + | <math>\lim_{x \rightarrow \inf}\frac{1}{x}</math> |
I'm going to assume that the answer (L) is 0. | I'm going to assume that the answer (L) is 0. | ||
− | Now, let me choose a super-duper small number, 0.001. Can I find another number ( | + | Now, let me choose a super-duper small number close to 0, 0.001, for <math>\epsilon \,</math>. Can I find another number (x) close to <math>\inf \,</math> that will give me <math>|\frac{1}{x} - 0| < 0.001 \,</math>? Of course, I can, and it would be something like 10,000, or anything bigger. |
− | It doesn't matter how close I get to 0, I can always find another number that will get me even closer. | + | It doesn't matter how close I get to 0, I can always find another number that will get me even closer. I never have to choose infinity itself, because that's not even a real number, but I can get closer and closer if I need to. |
So the limit is 0. | So the limit is 0. |
Revision as of 08:10, 14 April 2012
Limits
Definition of a limit:
<math>\lim_{x \rightarrow c} f(x) = L\,</math>
means that for all real values of <math>\epsilon > 0 \,</math>, there exists <math>\delta \,</math> for all <math>x \,</math> such that <math>0 < |x - c| < \delta \,</math>, we have <math>|f(x) - L| < \epsilon \,</math>.
In my own words:
Assume that the limit of <math>\lim_{x \rightarrow c} f(x) \,</math> is L.
Now, if we want to prove that it is L, we have to look at all possible numbers, especially small ones. Now, let's choose a number x that is awfully close to c, and see what f(c) gives us. It should get use well below this number we imagined.
Example
As a simple example, let's find:
<math>\lim_{x \rightarrow \inf}\frac{1}{x}</math>
I'm going to assume that the answer (L) is 0.
Now, let me choose a super-duper small number close to 0, 0.001, for <math>\epsilon \,</math>. Can I find another number (x) close to <math>\inf \,</math> that will give me <math>|\frac{1}{x} - 0| < 0.001 \,</math>? Of course, I can, and it would be something like 10,000, or anything bigger.
It doesn't matter how close I get to 0, I can always find another number that will get me even closer. I never have to choose infinity itself, because that's not even a real number, but I can get closer and closer if I need to.
So the limit is 0.
Derivatives
Integrals
Fundamental Theorem of Calculus
This is simply that the anti-derivative is the integral, as we've already said. This is one of those things that should make you go, "Wow!"