Understanding Limits: Find Out Why it Matters

[marciokoko]

I'm taking an online coursera course and the guy is explaining limits.

He doesn't really explain why he does this numerator thing.

 

[RUber]

You also didn't do a great job of explaining.
I will assume that the original question was something like:
find $\lim_{x\to 5} \frac{x^2 - 2x -15}{x-5}$

The reason he is doing "the numerator thing" is because the limit of a fraction that is only undefined at one point is the simplification of that fraction.
So, you need to know where your denominator is zero -- this is pretty clear.
How do you divide $\frac{x^2 - 2x -15}{x-5}$? There are many ways, but if you can factor the numerator, you might be able to use the easiest of them -- cancellation.
Since $x^2 - 2x -15 = (x-5)(x+3)$ you can simply cancel out the (x-5) for any place that the *original* function was defined i.e. $x\neq 5$.
Then you are left with a function that looks like (x+3) except it has an infinitely small hole in it at x=5. So...what is the limit as it approaches 5?

 

[HallsofIvy]

It's a bit hard to understand this because you have left out the first part of that. I think the problem is to find the limit, as x goes to 5 of
$$\frac{x^2- 2x- 15}{x- 5}$$

Now, polynomials, rational functions, etc. have the nice property that they are continuous wherever they are defined. That is, the limit, as x approaches that point, is the same as the value of the function at that point. So the first thing one would want to try is to evaluate the function there: This is a rational function but setting x= 5 gives $\frac{25- 10- 15}{5- 5}= \frac{0}{0}$. That is, of course, not a number because we cannot divide by 0- the function is NOT continuous at x= 5. However, there is a very important property of limits- the limit, as x goes to a, of f(x) depends upon values of f(x) close to a but does NOT depend on f(a), the value of f(x) at x= a. In more technical terms, "If there exist an interval, [b, c], containing a, such that f(x)= g(x) for all x in [b, c] except at x= a, then $\lim_{x\to a} f(x)= \lim_{x\to a} g(x)$

Here, $f(x)= \frac{x^2- 2x- 15}{x- 5}$. If we set x= 0 into that, we get "$\frac{0}{0}$" which is "undetermined". (Notice that that fact that $x^2- 2x- 15$ is 0 at x= 5 means that x- 5 is a factor of that polynomial. The reason he "does this numerator thing" is that he is showing that $x^2- 2x- 15= (x- 5)(x+ 3)$. So we can write $\frac{x^2- 2x- 15}{x- 5}= \frac{(x- 5)(x+ 3)}{x- 5}$. Again, that fraction is not defined at x= 5. But as long as x is not 5, $\frac{(x- 5)(x+ 3)}{x- 5}= x+ 3$ since the two "x- 5" terms cancel. So we have $\frac{(x- 5)(x+ 3)}{x- 5}= x+ 3$ for all x except x= 5 which means they have the same limit at x= 5. It is easy to see that the limit of x+ 3, as well as the original fraction, is 5+ 3= 8.

 

[marciokoko]

Sorry guys. I posted from my iphone and I had 2 images, the first containing the problem. I am guessing the second image replaced the first. :-)

Lucky for me you guys are sharp! :-)

Yes that was the original problem. And he did specify that f(x) is undefined at x = -5. He goes on to say that in this case, we can find a nicer function g(x) defined at x=-5 that agrees with f(x) near x=-5.

Finally he asks, find the lim of f(x) as x=-5.

What is not clear to me is what one thing has to do with the other? Iow, I understand we have that function that is undefined at x=-5. But what is this about finding a "nicer" function that agrees with f(x) near x=-5?

 

[RUber]

The nicer function makes it easier to see what the limit is as x gets close to -5.
By simplifying the original function to one that is easy to visualize, there is no guesswork.
You could always start plotting points near -5 by plugging them into your original function, but that can be time consuming and inaccurate.
What he is saying, as HallsofIvy stated above, is that we aren't looking for some arbitrary "nice" function, but one that looks exactly like the original function everywhere but the one point that it is undefined.
If such a function exists, and is continuous, then we know that the limit as x approaches -5 is simply the evaluation of the "nicer" function at x = -5.

 

[symbolipoint]

I'm taking an online coursera course and the guy is explaining limits.

He doesn't really explain why he does this numerator thing.

Agreed. Some of the explanation near the end of your image falls apart for me, but the method is clear. The function in original form, and in the simplified form is not permitted to accept x=5, but the LIMIT can still be found. However near to x=5 we choose, (x-5)/(x-5) will still be equal to 1 ; and so the limit will be (x+3)=(5+3)=8.