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- Thread starter ugu
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Hello. Only part (ii) instructs to discuss continuity. Are you asking for help at x=0, with the function in part (ii)?(Discuss the continuity of these functions)

[imath]\;[/imath]

I do not understand it myself but that was just the question: To discuss the continuity of these functions. It these was used, maybe there are more than one; if there are not, may be it was just error of typingHello. Only part (ii) instructs to discuss continuity. Are you asking for help at x=0, with the function in part (ii)?

[imath]\;[/imath]

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f(x) is said to be continuous at x=a if the following conditions are met.

1a) limit as x approaches a from the left of f(x) exists.

1b) limit as x approaches a from the right of f(x) exists.

2) The limits from 1a) and 1b) above are equal (say they both equal L)

3) f(a) = L

There are two functions and they want you to (separately) discuss the continuity at x=0.

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Hi. Just to confirm, which part are you currently working on?I do not understand it myself

Part (i)

Part (ii)

Part (iii)

When they say, "Discuss the continuity of the function", I think they mean discuss

What have you learned so far about function limits, as x

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No, that's not what I'd said.What you are trying to say is that, the statement is not clear regarding the [questions] to be [answered]?

I understand the exercise, but I don't like their wording "the continuity of the function" because that phrase implies that the function is continuous.

We cannot tell just by looking at the piecewise definitions in parts (ii) and (iii) whether either of those functions are continuous or not, at x=0. The reason why is because we haven't been given any values for the parameters m and a.

Perhaps there exists some value for m that makes the first function continuous. Maybe there's more than one value that works. Or, maybe no value of m will make the function continuous at x=0.

Likewise, the second function may or may not be continuous at x=0. If it can be, then there are one or more values of a that produce continuity.

You need to investigate and then discuss issues like these:

Whether each function can be continuous at x=0

If it can be continuous, then what value(s) of m or a are required

If it can not be made continuous, then state why not

No, that's not what I'd said.

I understand the exercise, but I don't like their wording "the continuity of the function" because that phrase implies that the function is continuous.

We cannot tell just by looking at the piecewise definitions in parts (ii) and (iii) whether either of those functions are continuous or not, at x=0. The reason why is because we haven't been given any values for the parameters m and a.

Perhaps there exists some value for m that makes the first function continuous. Maybe there's more than one value that works. Or, maybe no value of m will make the function continuous at x=0.

Likewise, the second function may or may not be continuous at x=0. If it can be, then there are one or more values of a that produce continuity.

You need to investigate and then discuss issues like these:

Whether each function can be continuous at x=0If it can be continuous, then what value(s) of m or a are requiredIf it can not be made continuous, then state why not

Recall that to test for continuity, you have to compute f(0),

But in this question, it has been given as ln a

Next limit x tends to zero of f(x) can be easily obtained by putting x=0 in the function...to get 0/1=0

Finally, you'll conclude that since f(0) is not equal to limit x tends to zero of f(x), then the function is discontinuous at x=0, regardless the value of a

========================================================

When you're asked to discuss the continuity... it mathematical means you should test the whether the function is continuous.

I put the value of a and m such way so that you can think just the way you've thought.

But if you look at the questions 2 and 3 well enough, no value of a or m can make the function continuous.

That's why I could easily generalise in the question 3 I solved

Unlike question 1, all you have to say is if k =lim x tends to 1 of f(x) then the function is continuous...I.e k=1/sqrt(2)

Sqrt means square root

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f(0) = ln(a) in part (iii)you have to compute f(0), But in this question, it has been given as ln a

f(0) = 1/m in part (ii)

Yes, 0/(-1)=0 in both parts (ii) and (iii). That tells us thelimit x tends to zero of f(x) can be easily obtained by putting x=0 in the function...to get 0/1=0

You're talking about part (iii). Your conclusion is incorrect. Thereconclude that since f(0) is not equal to [the limit,] then the function is discontinuous at x=0, regardless the value of a

I agree in part (ii) that no value of m will make 1/m equal zero. Therefore, the function in part (ii) cannot be made continuous.if you look at the questions 2 and 3 well enough, no value of a or m can make the function continuous.

Please look at part (iii) again.

yes, its over. thanksHi ugudansom. Were you able to finish part (iii)?