Differential and integral Calculus

Otis

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

[imath]\;[/imath]

ugu

Junior Member
Hello. Only part (ii) instructs to discuss continuity. Are you asking for help at x=0, with the function in part (ii)?

[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 typing

Jomo

Elite Member
You can't play the game unless you know the rules!!!!!

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.

Otis

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

Part (i)
Part (ii)
Part (iii)

When they say, "Discuss the continuity of the function", I think they mean discuss issues of continuity for those two functions. In other words, what conditions are needed, for each of those two functions to be continuous at x=0. Considering possible values of the constants m and a, is continuity at x=0 even possible? Those questions are discussion points.

What have you learned so far about function limits, as x approaches some fixed value? ugu

Junior Member
What you are trying to say is that, the statement is not clear regarding the question to be solved?

Otis

Elite Member
What you are trying to say is that, the statement is not clear regarding the [questions] to be [answered]?
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=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​ ugu

Junior Member
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=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​ RESPONSE

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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Otis

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

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

limit x tends to zero of f(x) can be easily obtained by putting x=0 in the function...to get 0/1=0
Yes, 0/(-1)=0 in both parts (ii) and (iii). That tells us the limit is zero. But, the limit doesn't tell us anything about the value of f(0). Remember, in a limit, x never takes on the value it approaches. In your limits, x gets as close to zero as you would like, closer and closer, but x will never equal 0.

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

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

Please look at part (iii) again. Otis

Elite Member
Hi ugudansom. Were you able to finish part (iii)? ugu

Junior Member
Hi ugudansom. Were you able to finish part (iii)? yes, its over. thanks

• Otis