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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.

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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#2

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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(Original post by

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)?

**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)?

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#5

(Original post by

i plugged it in but should i take the x <3 and x ≥3 into consideration?

**jan_ai**)i plugged it in but should i take the x <3 and x ≥3 into consideration?

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(Original post by

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.

**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.

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(Original post by

sorry typing error the 𝑥→3 is supposed to be below the "lim"

**jan_ai**)sorry typing error the 𝑥→3 is supposed to be below the "lim"

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#8

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)

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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#9

(Original post by

this is the question

**jan_ai**)this is the question

Anyway, this is a rather trivial task but it’s the way for more complicated examples.

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#10

(Original post by

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)

**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)

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#11

(Original post by

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.

**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.

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#12

(Original post by

idk about mathematical rigour, but maybe for A-level it is acceptable (?)

**the_cosmo_guy**)idk about mathematical rigour, but maybe for A-level it is acceptable (?)

Last edited by RDKGames; 3 weeks ago

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#13

(Original post by

i plugged it in but should i take the x <3 and x ≥3 into consideration?

**jan_ai**)i plugged it in but should i take the x <3 and x ≥3 into consideration?

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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#14

(Original post by

In your notes . / textbook I would expect you to have a definition of continuity and its requirements.

**davros**)In your notes . / textbook I would expect you to have a definition of continuity and its requirements.

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#15

(Original post by

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).

**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).

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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#16

(Original post by

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

**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

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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