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Author Topic: Example discussion - Limit Question by Michelle Jones  (Read 2554 times)
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« on: January 27, 2009, 11:21:10 PM »

I found this question and I was just wondering how you would explain the solution:

If LaTex \lim_{x \rightarrow 3} f(x) = 7, which of the following must be true?

I. f is continuous at x=3
II. f is differentiable at x =3.
III. f(3)=7

Thank you,
Michelle
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« Reply #1 on: January 27, 2009, 11:22:54 PM »

None of them have to be true.  You could easily have a function with f(3) undefined but LaTex  \lim_{x \rightarrow 3} f(x)=7.

Doug
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« Reply #2 on: January 27, 2009, 11:25:03 PM »

none must be true.  consider LaTex f(x) = \frac{{7(x-3)}^{2}}{(x-3)}

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« Reply #3 on: January 27, 2009, 11:26:35 PM »

My explanation would go something like this.  Information about the limiting behavior of a function as x-> a tells us absolutely nothing about the function's behavior *at* x = a.  The function could have a limit as
x-> a, but f(a) might not even exist, might exist and be different from the limit, or might exist and be the same as the limit.  In the latter case, we'd say the function is continuous at x=a.  And even if the limit as x-> a didn't exist, f(a) could exist and be any number, or not exist.  Function behavior at a point and limiting behavior of a function as x approaches the point are two completely different things. 

So the answer to the question is that none of those things need be true.

Mark Howell
Gonzaga High School
Washington, DC
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« Reply #4 on: January 27, 2009, 11:27:33 PM »

I.  In order for a function to be continuous at a point, the limit must
exist at a point and the function actually equal its limit.  (This is
something students need to know well for the exam).  So, for (I), we have
one of these two conditions (the limit) but we don't know anything about
the other (f(3) = 7?).  So doesn't have to be true.

II.  If the function doesn't have the be continuous, it sure doesn't have
to be differentiable.  Differentiable requires continuity.  Also,
functions with corners have limits but not derivatives at points, so
that's also a counter-example.  Definitely doesn't have to be true.

III.  Nope. The function can have a hole, for example.
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« Reply #5 on: January 27, 2009, 11:29:02 PM »

I found this question and I was just wondering how you would explain the solution:

If LaTex \lim_{x \rightarrow 3} f(x) = 7, which of the following must be true?

I. f is continuous at x=3
II. f is differentiable at x =3.
III. f(3)=7

Thank you,
Michelle


Well, the way I would explain it is to first note
that, in order of strength (something I'd want to
make sure the student knows, so I'd go to the trouble),
we have

II implies I implies III.

Thus, an efficient strategy (something I'd also want
a student to know) would be to first see if III can
fail. Since III is the weakest, if III fails, then
the other two will fail for the same example. If you
can't get III to fail, then see if I can fail, because
II will also fail for the same example. If you can't
get I to fail, then see if you can get II to fail.

Doing this will give you a lower bound on what can
fail. [Whatever example(s) you have in hand, if any,
will definitely show things that don't have to be
true. However, there might be more things that don't
have to be true than you were able to uncover.] To
get an upper bound on what can fail (which hopefully
will be the same as the lower bound you got, so that
you can answer the question), go back and see if any
of the reverse implications can be established with
what is given. Of course, if all three fail for your
example, then this step isn't needed.

As to the particulars at hand in the problem you
gave, it's of course easy to get an example where
III fails. [Just consider a function whose value at
x = 3 isn't equal to its limit at x = 3. For example,
f(x) = 7 for all x not equal to 3 and f(3) = 8.]

Dave L. Renfro

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« Reply #6 on: January 27, 2009, 11:29:58 PM »

Ask your students to make the a sketch that fits this description.  Make
a filled in circle at (0,7) and an open circle at (3,7), then connect
these circles with a piece of a horizontal line.  Now make a filled in
circle at (6,7) and connect this circle to the open circle at (3,7) with
another piece of a horizontal line.  Finally, add to this sketch a
filled in circle at (3,6).  They will have graphed a function with
domain [0,6].  Call the function f.  Ask your whether it is true that
the limit of f at 3 is 7? Then ask how any of the three statements in
your list are true about function f?

Sincerely,

Richard Sisley
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« Reply #7 on: January 27, 2009, 11:30:59 PM »

None must be true.
 
If II is true then I and III are also.
If III is true then I is true.and conversely.
 
This could be a new style question. Which must be true to  make the others true also?
 
 
Lin.

Lin McMullin
Richardson, Texas

www.LinMcMullin.net
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