# Differential = continuity theorem

1. Mar 4, 2013

2. Mar 4, 2013

### MostlyHarmless

One method we have for solving limits is to simply take the the value that our variable is approaching and plug it in for our variable. In this case we have the limit as x approaches x0 of f(x)-f(x0). So by taking x0 and plugging it in for x, we get f(x0)-f(x0)=0. Therefor the limit(x-->x0) f(x)-f(x0)=0

Hope this helps. Let me know if you need any clarification.

3. Mar 4, 2013

### jimmyly

Ohhh okay thanks for the response. So is it basically the same as "simple" limits like lim(x-->2) x+1 where you just plug 2 into x?

4. Mar 4, 2013

### MostlyHarmless

Yep, that should be your goto when trying to solve a limit. And if that doesnt work, well, there are other methods, but you'll get to those if you havent already.

5. Mar 4, 2013

### jimmyly

thanks a lot!

6. Mar 4, 2013

### HallsofIvy

Staff Emeritus
Be careful here. A lot of students get the impression that "limit" is just a fancy way of talking about the value of a function. That is often true because so many of the functions we use are 'continuous' which just means that $$\lim_{x\to a} f(x)= f(a)$$. The fact is that "almost all" functions are NOT continuous- we use such functions so often just because the are "nice".

For example, suppose f(x) is defined by f(x)= x for x> 0, f(x)= 3x for x< 0 and f(0)= 4. Then f(0)= 4, of course, but $$\lim_{x\to 0} f(x)= 0$$.

In this particular example, $\lim_{x\to x_0} x- x_0= 0$ because, given any number $\epsilon> 0$, if we take $\delta= \epsilon$, $|x- x_0|< \delta$ then $|f(x)- 0|< \epsilon$ simply because $f(x)= x- x_0$.
Typically, what we do is prove the "trivial limits", $\lim_{x\to a} x= a$ and $\lim_{x\to a} C= C$, for C any constant, then prove the theorems, "if $\lim_{x\to a} f(x)= F$, and $\lim_{x\to a} g(x)= G$, then $\lim_{x\to a} f(x)+ g(x)= F+ G$ and $\lim_{x\to a} f(x)g(x)= FG$" to show that any polynomial is continuous for all x.

7. Mar 4, 2013

### jimmyly

Hello HallsoIvy,

Thank you for your response. I just receive quick calculus by taylor and it has what you are explaining at the end of your post. I don't quite understand it at the moment I just started self-studying about a week and a half ago. Can you suggest some resources that may help me understand?

I know of khan academy and mit ocw.

EDIT: I understand what you are explaining up until "typically, what we do is prove the..."

8. Mar 4, 2013

### MostlyHarmless

Just to be sure, I don't want to go around tossing out inaccurate advice. My explanation is correct and does suffice for this instance, right? As the instructor just simply wrote down the limit=0 on the board, I just felt like a simple "this is where that came from" answer was all that was needed.

9. Mar 4, 2013

### jimmyly

I understand it now. you are both very helpful. thank you again!

10. Mar 5, 2013

### micromass

Staff Emeritus
Sure, but you should have mentioned somewhere that it only holds for functions f that are continuous. The method certainly does not work for all possible functions f. So I guess that the important part is that f(x)=x is continuous and that the method therefore works.

11. Mar 5, 2013

### HallsofIvy

Staff Emeritus
Yes, for this particular problem what you said was correct. I just wanted to make sure that no one got the impression that we find limits, in general, by simply evaluating the function!

12. Mar 5, 2013

### HallsofIvy

Staff Emeritus
I assume that you have seen the basic definition of "limit":
$\lim_{x\to a} f(x)= L$ if and only if, given $\epsilon> 0$, there exist $\delta> 0$ such that is $|x- a|< \delta$ then $|f(x)- L|<\epsilon$.

In particular, if f(x)= C, a constant, then $|f(x)- C|= |C- C|= 0$ which is always less than any positive $\epsilon$ so $\lim_{x\to a} C= C$. If f(x)= x, then $|f(x)- a|= |x- a|< \epsilon$ so it is enough to take $\delta= \epsilon$.

If $\lim_{x\to a} f(x)= L$, it follows, as above, that given $\epsilon> 0$, there exist $\delta> 0$ such that if [/itex]|x- a|< \delta[/itex] then $|f(x)- L|< \epsilon$. But in that case, |C||f(x)- L|= |Cf(x)- CL|< \epsilon[/itex] also, so
$\lim_{x\to a} Cf(x)= CL$.

If $\lim_{x\to a} f(x)= F$ and $\lim_{x\to a} g(x)= G$, then, given $\epsilon> 0$ there exist $\delta_1> 0$ such that if $|x- a|< \delta_1$ then $|f(x)- F|< \epsilon/2$ and $\delta_2> 0$ such that if $|x- a|< \delta_2$ then $|g(x)- G|< \epsilon$.
($\delta_1$ and $\delta_2$ are not necessarily the same- we will handle that in a minute. Also note the "$\epsilon/2$". Since $\epsilon$ could be any positive number, and $\epsilon/2$ is also positive, we can use that as well. You will see why we want $\epsilon/2$.)

If we define $\delta$ to be the smaller of $\delta_1$ and $\delta_2$, then if $|x- a|< \delta$ we have both $|x- a|< \delta_1$ and $|x- a|< \delta_2$, so that $|f(x)+ g(x)- (F- G)|= |f(x)- F+ g(x)- G|\le |f(x)- F|+ |g(x)- G|<\epsilon/2+ \epsilon/2= \epsilon$. That is, if $\lim_{x\to a} f(x)= F$, $\lim_{x\to a} g(x)= G, then [itex]\lim_{x\to a} f(x)+ g(x)= F+ G$.

Those proofs are in any Calculus text.

13. Mar 5, 2013

### MostlyHarmless

Ok, thank you for the clarification. When you're just now learning something, less is more I think. I tend to get bogged down by proofs and theorems when learning new material. It's easier for me to understand the process "numerically" and be able to get an answer, and then go back and see how this is true symbolically.

14. Mar 9, 2013

### HallsofIvy

Staff Emeritus
Note, further, that "continuous" is defined as having the property that $\lim_{x\to a} f(x)= f(a)$. In fact, "almost all" functions are not continuous at any point. What is true is that the property of being continuous is so useful that almost all of the functions we use are continuous.