A limit stumper
Stephen J. Herschkorn
assigned by his professor: Determine the limit of [sqrt(6-x) - 2] /
[sqrt(3-x) - 1] as x approaches 2. Here's the catch: The problem is
from early in the book. Limits have been discussed, but the derivative
(let alone l'Hopital) has not been introduced yet. I don't see how to
do it wthout a notion of derivatives. Do you?
(I do see how to do it with only elementary facts about derivatives.)
--
Stephen J. Herschkorn sjher...@netscape.net
The World Wide Wade
Why not use the old standby: Multiply top by [sqrt(6-x) + 2] /
[sqrt(6-x) + 2], multiply the bottom by [sqrt(3-x) + 1] /
[sqrt(3-x) + 1].
Stephen J. Herschkorn
Stephen J. Herschkorn
That approach, which was my first thought, does not solve the problem:
You still have 0 in both numerator and denominator.
Dave L. Renfro
>> Why not use the old standby: Multiply top by [sqrt(6-x) + 2] /
>> [sqrt(6-x) + 2], multiply the bottom by [sqrt(3-x) + 1] /
>> [sqrt(3-x) + 1].
Stephen J. Herschkorn wrote:
> That approach, which was my first thought, does not solve
> the problem: You still have 0 in both numerator
> and denominator.
Multiply numerator and denominator by both conjugates,
the conjugate of the numerator and the conjugate of the
denominator, and then cancel the common (2-x) factor that
appears. A similar example was posted in alt.math.undergrad
about a year ago, by the way. When I worked your limit
problem just now, I was reminded of seeing something
like this recently, so I searched my posts for
"rationalize+numerator-and-denominator".
http://groups.google.com/group/alt.math.undergrad/msg/27161334c7c40271
Dave L. Renfro
Pouya D. Tafti
First multiply by
sqrt(3-x) + 1
------------- ;
sqrt(3-x) + 1
you get 2-x in the denominator which you can rewrite as
(sqrt(6-x) - 2) x (sqrt(6-x) + 2)
to get
(sqrt(6-x) - 2) x (sqrt(3-x) + 1) sqrt(3-x) + 1
--------------------------------- = ------------- .
(sqrt(6-x) + 2) x (sqrt(6-x) + 2) sqrt(6-x) + 2
The limit as x->2 will be 1+1 / 2+2 = 1/2.
--
Pouya D. Tafti
p dot d dot tafti at ieee dot org
Stephen J. Herschkorn
Dave L. Renfro wrote:
>The World Wide Wade wrote:
>
>
>
>>>Why not use the old standby: Multiply top by [sqrt(6-x) + 2] /
>>>[sqrt(6-x) + 2], multiply the bottom by [sqrt(3-x) + 1] /
>>>[sqrt(3-x) + 1].
>>>
>>>
>
>Stephen J. Herschkorn wrote:
>
>
>
>>That approach, which was my first thought, does not solve
>>the problem: You still have 0 in both numerator
>>and denominator.
>>
>>
>
>Multiply numerator and denominator by both conjugates,
>the conjugate of the numerator and the conjugate of the
>denominator, and then cancel the common (2-x) factor that
>appears.
>
Very good. I wish I had seen that.
BTW, with derivatives but without l'Hopital, I set u = sqrt(3 - x). We
then rewrite the expression as [f(u) - f(1)] / (u - 1), and we can
easily determine f'(1).
The World Wide Wade
"Stephen J. Herschkorn" <sjher...@netscape.net> wrote:
> [Apologies for the blank message I may have posted]
>
> The World Wide Wade wrote:
>
> >In article <4554DC1A...@netscape.net>,
> > "Stephen J. Herschkorn" <sjher...@netscape.net> wrote:
> >
> >
> >
> >>I was helping a student with calculus, and he posed me this problem
> >>assigned by his professor: Determine the limit of [sqrt(6-x) - 2] /
> >>[sqrt(3-x) - 1] as x approaches 2. Here's the catch: The problem is
> >>from early in the book. Limits have been discussed, but the derivative
> >>(let alone l'Hopital) has not been introduced yet. I don't see how to
> >>do it wthout a notion of derivatives. Do you?
> >>
> >>(I do see how to do it with only elementary facts about derivatives.)
> >>
> >>
> >
> >Why not use the old standby: Multiply top by [sqrt(6-x) + 2] /
> >[sqrt(6-x) + 2], multiply the bottom by [sqrt(3-x) + 1] /
> >[sqrt(3-x) + 1].
> >
> >
>
> That approach, which was my first thought, does not solve the problem:
> You still have 0 in both numerator and denominator.
Cancel the 2-x terms!
Gib Bogle
Write x = 2 + d, expand the Taylor series for the sqrts, drop the terms
in d^2 and higher powers.
A N Niel
<bo...@ihug.too.much.spam.co.nz> wrote:
Someone doesn't know derivatives, but does know Taylor series?
Kira Yamato
>
> Someone doesn't know derivatives, but does know Taylor series?
It could be done.
In Heinrich Dorrie's book "100 Great Problems of Elementary
Mathematics" it shows how Newton had used a series of inequalities and
some known algebraic identities (no derivatives) to derive power series
for the exponential function, the logarithm, and the trignometric
functions.
--
-kira
Gib Bogle
I guess I should have said power series. But you may be right, one
probably learns calculus before learning the power series for sqrt(1-x).
I can't remember which came first in my education.
Herman Rubin
Use the standard device; let x = 2 + y. Then the expression becomes
[sqrt(4-y) - sqrt(4)]/[(sqrt(1-y) - sqrt(1)], and using
sqrt(a-y)-sqrt(a) = -a/[(sqrt(a-y)+sqrt(a)],
the result is immediate.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Brian
Can't you just use a calculator and see where it leads?
Using x = 2.1 the expression results in 0.490
Using x = 2.001 the expression results in 0.49990
I didn't even bother with 2.00001, it's pretty clear this
will end up as 0.49999999....(repeating indefinitely).... = 0.5
Christopher Night
What do you get for the limit as x goes to 8 of:
x^2 + ln((x-8)^2) / x^8
Stan Liou
The identity Sum[ C(2n,n) x^n ] = 1/sqrt[1-4x] can be shown using
combinatorial counting arguments involing the binomial coefficient,
with no reference to the derivative. Of course, that is not
something a beginning student of calculus would be expected to know,
but nevertheless it is possible.
---
Stan Liou
Zdislav V. Kovarik
On Tue, 14 Nov 2006, Brian wrote:
Experimentig with calculator can be tricky. Try the limit as x goes to 0
of
1/(sin(x))^2 - 1/x^2.
Another infamous example is the limit of the recursively defined sequence
a(0) = 1 - 1/e
a(n+1) = 1 - (n+1)*a(n)
which should converge to 0 rather slowly, but calculator results will
diverge rather quickly after first few steps.
Clue: a(n) is the integral from 0 to 1 of x^n * e^(x-1).
Now to the original problem:
(sqrt(6-x)-2)/(sqrt(3-x)-1) can be transformed using rationalization into
(sqrt(3-x)+1)/(sqrt(6-x)+2).
Check it, and find the limit (as x goes to 2) without difficulty.
Cheers, ZVK(Slavek).
Herman Rubin
This is an indication, but not a proof. Also, the closer
x is to 2, the less accuracy there is in the result.