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marciokoko

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- #1

marciokoko

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- #2

RUber

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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?

- #3

HallsofIvy

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[tex]\frac{x^2- 2x- 15}{x- 5}[/tex]

Now, polynomials, rational functions, etc. have the nice property that they are

Here, [itex]f(x)= \frac{x^2- 2x- 15}{x- 5}[/itex]. If we set x= 0 into that, we get "[itex]\frac{0}{0}[/itex]" which is "undetermined". (Notice that that fact that [itex]x^2- 2x- 15[/itex] is 0 at x= 5

- #4

marciokoko

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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?

- #5

RUber

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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.

- #6

symbolipoint

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

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

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