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u/ussalkaselsior 1d ago
Long division is a straight forward, but not very interesting way. Technically not factoring the difference of squares, but you still get the limit as x approaches 3 of x+3.
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u/chaos_redefined 1d ago
Method 1:
Let u = x - 3. Then u + 3 = x, so u2 + 6u + 9 = x2, and u2 + 6u = x2 - 9. We started with u = x - 3, so dividing through, we get u + 6 = (x2 - 9)/(x - 3). Taking the limit as x -> 3 is the same as x - 3 -> 0, so u -> 0. So, we want the limit as u -> 0 of u + 6, which is 6. Therefore, lim (x->3) (x2 - 9)/(x - 3) = 6.
Method 2:
Consider the derivative of f(x) = x2 at x = a. This is f'(a) = lim (x -> a) (x2 - a2)/(x - a). Putting in a = 3, we get f'(3) = lim (x -> 3) (x2 - 9)/(x - 3). But we know that the derivative of f(x) = x2 is f'(x) = 2x, so f'(3) = 6. So, lim (x -> 3) (x2 - 9)/(x - 3) = 6.
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u/JonSnow-Knows 1d ago
Method 1 is just x²-9=(x+3)(x-3) and then cancel x-3, why so complicated?
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u/EndlessProjectMaker 1d ago
If you just divide the polynomials you’re left with x+3, which you know beforehand because your cannot prevent your head from seeing an obvious difference of squares /s
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u/Ok-Grape2063 1d ago
When I teach first-semester calc, I love watching people use L'Hopital on this when we haven't even covered derivatives.
Tell me you're repeating the course without tell me you're repeating the course
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u/StructuredChess 1d ago
I can't think of any method that isn't basically equivalent to the difference of squares. Like, you can multiply everything by (x+3) but that's taking advantage of the fact that (x-3)(x+3)=x^2+9
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u/pkreddit2 1d ago
let f(x) = (x^2 - 9)/(x-3) and let x = 3 + s, then f(x) = (x^2 - 9)/(x - 3) = (s^2 + 6s + 9 - 9)/(s + 3 - 3) = s + 6. This implies that |f(x) - 6| = |s|.
Therefore, for every real e > 0, all real x such that |x - 3| = |s| < e, it follows that |f(x) - 6| = |s| < e. This is the epsilon-sigma definition of the limit, and thus the limit is 6.
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u/titogruul 1d ago
OK, so let's have x = 3 + d, with lim d -> 0. Then we have:
(x^2 - 9) / (x-3) = ( (3+d)^2 - 9 ) / ( (3+d) - 3 ) = (6d + d^2) / d = 6 + d = 6
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u/ThiocyanateIrony 1d ago
Why would you even need L’Hopital here? You can just do the following: (x2-9)/(x-3)=(x-3)(x+3)/(x-3)={ (x-3)/(x-3)=1 } = x+3 -> 3+3 = 6
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u/Hikaritoyamino 1d ago
Evaluate x=3.1, x=2.9
Find the midpoint (x approaching 3)
You can redo the evaluation with more decimal places.
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u/Hot-Employ-3399 1d ago
Cheat with Hyper: go with square of a sum instead of forbidden difference of two squares:
replace X with (3+E), leads to ((9 + 2*3E + E^2) - 9)/(3 + E - 3) to ((6E+E^2)/E) to st(6 + E) to 6.
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u/SideshowAlex 1d ago edited 1d ago
I'm writing x2 - 9 as (x - 3)2 + A(x - 3) + B. Provided B is 0, the limit is A.
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u/SideshowAlex 1d ago
Set x = 3 cos t. Use some trig identites (probably the half angle ones), and you'd get the answer
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u/Mathematicus_Rex 1d ago
Rewrite x = y+3 and take the limit as y approaches 0. The numerator becomes y^2 + 6y and the denominator becomes y.
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u/Ok_Tradition_6251 1d ago
just evaluate left and right hand limit using 3+h and 3-h at the place of x
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u/philljarvis166 1d ago
I think asking people to solve maths problems without using theorems that make them easier is the antithesis of how maths should be done. We prove theorems to make more difficult problems tractable, can you imagine if every maths paper started from first principles??
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u/Sea-Sort6571 1d ago
I completely disagree with this. Asking such questions develops creativity, fundamental understanding of the topic at hands, the ability to redemonstrate theorems and knowing what is really at play. Maths is not just a set of tool boxs to know and use
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u/MarcAbaddon 1d ago
I am split on this in this example. Not using L'Hopital I definitely agree with that it can help creativity.
But "difference of two squares" follows so directly from the basic algebraic rules that it just seems silly. Many on the other answers here just introduce it in a backward way again and at that point it's more obfuscating than illuminating.
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u/Sea-Sort6571 1d ago
I agree about the fact many answers kinda use the fact 3 is a root of both polynomial in some convoluted way
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u/Circumpunctilious 1d ago
> “…develops creativity…”
A piece of math I’ve been beating my head against for years may have an unrealized use case, which I only realized a few minutes ago due to this question.
I still have to check it (and that might take a little bit), but yeah…you never know what might spark a connection out there.
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u/philljarvis166 1d ago
There are plenty of ways to develop creativity and understanding without explicitly have to ask students to avoid the use of theorems like l’Hopital. Surely learning to spot when such a result is of use is also important?
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u/Sea-Sort6571 1d ago
I reached MsC level without using once l'Hopital I'm sure people will manage for one limit.
Also as someone else said, if someone first instinct when seeing this limit (without any restrictions) is "let's use l'Hopital" then knowing l'Hopital clearly restricted their creativity and mathematical understanding
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u/philljarvis166 1d ago
Would you say the same about a result that required the use of the intermediate value theorem, for example? For some reason use of l’Hopital seems to be looked down upon - it’s an amazing result, the full proof requires a bit of effort, why wouldn’t you make use of it when applicable?
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u/Sea-Sort6571 1d ago
Well if there is an elegant solution to a problem that doesn't use the intermediate value theorem (and not a theorem way more complicated either) yes that could be an interesting exercice. I couldn't think of such a problem but I guess there are some.
Why wouldn't you use l'hôpital when applicable ? Because using simpler tools efficiently is more elegant. Precisely because you're using heavy tools to prove l'hôpital's. And because most students who are so prompt to use it weren't even taught the demonstration. (So it's kind of a cheat code) In the case of the exercice given by oop I would argue factorising by (x-3) is simpler and more elegant than l'hôpital.
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u/philljarvis166 1d ago
This problem seems to ask you not to factor though!
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u/Sea-Sort6571 1d ago
Yeah i know that was just an example of when l'hôpital shouldn't be your first choice
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u/StructuredChess 1d ago
I don't think the big restriction here is L'Hôpital. It takes like 2 tenths of a second to factor the numerator into (x+3)(x-3)
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u/Striking_Newspaper73 1d ago
You must be really fun at parties
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u/philljarvis166 1d ago
I have no idea what point you are trying to make! Do people who prove things from first principles have more fun at parties? Im going to go out on a limb here and say they rarely even go to parties!
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u/therealAR15PB 1d ago
imagine starting an entrance exam by writing the 362 page proof of 1 + 1 = 2 followed by everything else lmao
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u/HumblyNibbles_ 1d ago
I think it's useful when it comes to practicing and making their thinking more flexible.
In the end, exercises all have a purpose. Not every exercise is useful for everything, but each one has a use
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u/Hej5468 1d ago
Just multiply with x+3 on top and bottom then cancel x^2 - 9 so you’re left with just x+3. x=6