jan_ai
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The function 𝑓 on ℝ is defined by
f(x)= {x^2-2x+3}, x <3
= {4x-6}, x ≥3
(i) Find lim 𝑓(𝑥)
𝑥→3
(ii) Determine whether 𝑓(𝑥) is continuous at 𝑥 = 3.Give a reason for your answer.
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ThomH97
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Do you have a guess for i) that you can try to verify (or you can just plug it in if you don't need to be that rigorous)?
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DFranklin
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Where has this question come from? (It's not a calculus question, incidentally).
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jan_ai
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(Original post by ThomH97)
Do you have a guess for i) that you can try to verify (or you can just plug it in if you don't need to be that rigorous)?
i plugged it in but should i take the x <3 and x ≥3 into consideration?
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ThomH97
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(Original post by jan_ai)
i plugged it in but should i take the x <3 and x ≥3 into consideration?
I had assumed there was some formatting issue with your limit arrow so you didn't specify whether the limit was from above or below. If the question hasn't requested one or the other, then I would state both, specifying which is which.
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jan_ai
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(Original post by ThomH97)
I had assumed there was some formatting issue with your limit arrow so you didn't specify whether the limit was from above or below. If the question hasn't requested one or the other, then I would state both, specifying which is which.
sorry typing error the 𝑥→3 is supposed to be below the "lim"
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jan_ai
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(Original post by jan_ai)
sorry typing error the 𝑥→3 is supposed to be below the "lim"
this is the question
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the_cosmo_guy
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i) it looks that you need to find the limit from both directions
when x approaches 3 from below, it's the first case, so it's 9-6+3=6
when x approaches 3 from above, it's the second case, so it's 12-6=6
ii) it is continous as both limits coincides, so there is not gap/discontinuity in the function (you can see this by plotting the function)
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RDKGames
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(Original post by jan_ai)
this is the question
Check the limit from both sides. If these agree, this is the limit as x tends to 0.

Anyway, this is a rather trivial task but it’s the way for more complicated examples.
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RDKGames
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(Original post by the_cosmo_guy)
i) it looks that you need to find the limit from both directions
when x approaches 3 from below, it's the first case, so it's 9-6+3=6
when x approaches 3 from above, it's the second case, so it's 12-6=6
ii) it is continous as both limits coincides, so there is not gap/discontinuity in the function (you can see this by plotting the function)
ii) it is not sufficient to say that the existence of the limit implies continuity at that point. I would leave it to OP to figure out the requirement for continuity.
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the_cosmo_guy
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(Original post by RDKGames)
ii) it is not sufficient to say that the existence of the limit implies continuity at that point. I would leave it to OP to figure out the requirement for continuity.
idk about mathematical rigour, but maybe for A-level it is acceptable (?)
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RDKGames
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(Original post by the_cosmo_guy)
idk about mathematical rigour, but maybe for A-level it is acceptable (?)
A-level doesn’t concern the notion of continuity AFAIK.
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davros
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(Original post by jan_ai)
i plugged it in but should i take the x <3 and x ≥3 into consideration?
In your notes . / textbook I would expect you to have a definition of continuity and its requirements.

These conditions have been hinted at above, but are you clear on what you need to show in order to check for continuity?
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DFranklin
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(Original post by davros)
In your notes . / textbook I would expect you to have a definition of continuity and its requirements.
Not to be cynical, but I'm expecting the OP to have no such definition. It seems an oddly phrased question to be asking if the question writer properly understands continuity. (Not *impossible*, but odd).
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davros
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(Original post by DFranklin)
Not to be cynical, but I'm expecting the OP to have no such definition. It seems an oddly phrased question to be asking if the question writer properly understands continuity. (Not *impossible*, but odd).
Re-reading the question, I wonder if a rigorous approach is needed / expected, or whether they just want the student to be able to comment that both "branches" of f(x) are individually continuous, and as their separate limits tend to f(3), then that's all that's needed.

But hard to tell, as standards vary enormously across unis from what I can see, and this may be an intermediate test, or an end-of-year exam question, or something else entirely
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DFranklin
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(Original post by davros)
Re-reading the question, I wonder if a rigorous approach is needed / expected, or whether they just want the student to be able to comment that both "branches" of f(x) are individually continuous, and as their separate limits tend to f(3), then that's all that's needed.

But hard to tell, as standards vary enormously across unis from what I can see, and this may be an intermediate test, or an end-of-year exam question, or something else entirely
It's hard to tell; my (somewhat uncharitable) guess is that this falls under the umbrella of "someone not really understanding limits properly themselves" setting a question involving limits.

To my mind, part (i) is quite difficult to answer "properly". Either you bash into a somewhat non-trivial epsilon-delta proof, or you need to justify that both partial definitions are continuous, and therefore that their limits tend to the obvious values, and as part of that I think you really need to do some level of justifying extending the domain of the LHS definition to include x = 3 (i.e. something like g(x) = x^2-2x+3 is cts on R, so since f = g for x < 3, the limit of f as x->3- is g(3)).

But it's also not that clear how to answer it "improperly" either: you still run into the problem that the "x^2-2x+3" definition doesn't actually run to x=3, so although you'd very much *like* to just "stick x = 3", I'm not sure you can without writing something patently false.

There's also the fact that (i) tacitly assumes the limit exists; I think an analysis question would generally say something like "show lim f(x) exists and find its value",

And then of course (ii) is completely trivial if aimed at people who have even a basic understanding of limits / continuity; not impossible (as a trivial 1 mark rider), but a bit odd.

But it's all just speculation, of course.
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