Limit of a function-can't solve few problems
boris
I'm trying to find limits of a functions but some
problems are giving me a hard time
Some are easy to solve: find limit of a function
(3*x^2 + 13*x + 4)/(x^3 + 4*x^2 + x + 4) as x-->(-4)
First I find common zeros of both polynomials and then
cancel them,and then I plug in -4 into equation and get
the limit->(-11/17)
But the following I was not able to solve:
lim(x^3 + x) / (x^5 - x) as x-->0
lim(x - 1) / ( sqrt(x) - 1 ) as x-->1
lim(sqrt(x+2)- sqrt(2)) /(x) as x-->0
Well there are a couple more I couldn't solve, but I'm
hoping if I get to understand these I will also be able to
solve others
Rational function has polynomial both in numerator and
denominator,and common zeros of the two must be
canceled out. Take the following function for example:
(3*x^2 + 13*x + 4)/(x^3 + 4*x^2 + x + 4) as x-->(-4)
If I'd just plug in -4 into this equation I would get 0/0,
which I think is called indeterminate form.So I first have
to cancel common zeros out.
But the thing is, that after I do that it is no longer the
same function and so the limit also shouldn't be the same.
Afterall limits for both polynomials(before I cancel zeros
out) in numerator and denominator are zero,and I
understand you can't divide zero by zero,but still the
fact remains that original function is not(or by my
reasoning shouldn't be) the same as rational function we
get after we cancel out common zeros.
I know some will argue that fraction doesn't change if we multiply by same number both numerator and denominator ...
thank you
Brian M. Scott
<failu...@yahoo.co.uk> wrote in
<news:4797604.11347760123...@nitrogen.mathforum.org>
in alt.math.undergrad:
[...]
> But the following I was not able to solve:
> lim(x^3 + x) / (x^5 - x) as x-->0
Factor the numerator and denominator and cancel the common
factors; there's one common factor that should be completely
obvious. (It's actually the only one that you need to find,
though in fact the whole numerator cancels.)
> lim(x - 1) / ( sqrt(x) - 1 ) as x-->1
x - 1 = sqrt(x)^2 - 1^2; now factor it.
> lim(sqrt(x+2)- sqrt(2)) /(x) as x-->0
Do you know the procedure for rationalizing a denominator?
For example,
3 / (2 - sqrt(3)) =
(3 / (2 - sqrt(3)) * (2 + sqrt(3)) / (2 + sqrt(3)) =
(6 + 2 sqrt(3)) / (4 - 3) =
6 + 2 sqrt(3).
A similar procedure works here: multiply the fraction by
(sqrt(x+2) + sqrt(2)) / (sqrt(x+2) + sqrt(2)),
then simplify.
Brian
Rob Johnson
boris <failu...@yahoo.co.uk> wrote:
>Rational function has polynomial both in numerator and
>denominator,and common zeros of the two must be
>canceled out. Take the following function for example:
>
>(3*x^2 + 13*x + 4)/(x^3 + 4*x^2 + x + 4) as x-->(-4)
>
>If I'd just plug in -4 into this equation I would get 0/0,
>which I think is called indeterminate form.So I first have
>to cancel common zeros out.
>
>But the thing is, that after I do that it is no longer the
>same function and so the limit also shouldn't be the same.
>Afterall limits for both polynomials(before I cancel zeros
>out) in numerator and denominator are zero,and I
> understand you can't divide zero by zero,but still the
>fact remains that original function is not(or by my
>reasoning shouldn't be) the same as rational function we
>get after we cancel out common zeros.
>I know some will argue that fraction doesn't change if we
>multiply by same number both numerator and denominator ...
Since x = -4 makes both numerator and denominator 0, both must have x+4
as a factor. So factoring, we get
3x^2 + 13x + 4 (x + 4)(3x + 1)
------------------ = ----------------
x^3 + 4x^2 + x + 4 (x + 4)(x^2 + 1)
Now (x+4)/(x+4) is the function 1 everywhere but at x = -4 where the
former is not defined. However, when taking limits, we are not at
all concerned about the value of the function at the limit point, but
the values at points arbitrarily near the limit point. So it doesn't
matter that the functions differ at x = -4. Therefore, we have
x + 4
lim ----- = 1
x->-4 x + 4
Since for all x near -4, (x + 4)/(x + 4) is actually equal to 1.
Thus, for all x near -4,
3x^2 + 13x + 4 3x + 1
------------------ = -------
x^3 + 4x^2 + x + 4 x^2 + 1
So the limit of both sides as x tends to -4 is the same, even though
the value of the left hand side is undefined at x = -4.
The function (3x + 1)/(x^2 + 1) is continuous at x = -4, so you can
simply plug in x = -4 to get the limit.
Of course, another approach which works is to use L'hospital's rule,
but that requires calculus, which is not needed for this problem.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
boris
boris
lim(sqrt(3x+1) - 2)/(sqrt(x) - 1) as x-->1
and
lim(x - sqrt(4x))/(sqrt(x) - 2) as x-->4
I did multiply both the numerator and denominator with
(sqrt(3x+1) + 2) in the first and with (x + sqrt(4x) in
the second case but still couldn't find anything to cancel
out
Dave L. Renfro
Hey, these are neat! Try rationalizing both the numerator and the
denominator. For example, with the first one, multiply both the
numerator and denominator by [sqrt(3x+1) + 2][sqrt(x) + 1].
Dave L. Renfro
Dave L. Renfro
This is a very legitimate question.
Yes, it's no longer the same function after canceling the
factors of x+4, but the functions are exactly the same
for all (nearby) values of x not equal to -4, and hence
their limits for x --> -4 will be the same.
Consider the graphs of y = x+2 and y = (x^2 - 4)/(x-2).
The first one will be a line and the second one will be
the same line with a hole (a single point missing). They'll
look like the following, except they'll be at a 45 degree
angle instead of horizontal:
----------------------------------------------------
--------------------o-------------------------------
Note that there is no difference between approaching the point
where the hole is in the second line than approaching the
corresponding point (which now doesn't have a hole) in the
first line. Remember, the definition of a limit involves
behavior *near the point*, not *at the point*, so if two
functions have the exact same behavior near the point,
their limiting behavior as you approach the point will
be the same.
Dave L. Renfro
Paul Sperry
<32459054.1134857866...@nitrogen.mathforum.org>, boris
<failu...@yahoo.co.uk> wrote:
> Well there are still two I can't solve
>
> lim(sqrt(3x+1) - 2)/(sqrt(x) - 1) as x-->1
>
> and
>
> lim(x - sqrt(4x))/(sqrt(x) - 2) as x-->4
[...]
Dave Renfro has given you an excellent hint for the first one. For the
second, note x - sqrt(4x) = sqrt(x)*(sqrt(x) - 2).
--
Paul Sperry
Columbia, SC (USA)
Darrell
"boris" <failu...@yahoo.co.uk> wrote in message
news:29547060.1134857941...@nitrogen.mathforum.org...
> Could someone also answer the following?
>
> Rational function has polynomial both in numerator and
> denominator,and common zeros of the two must be
> canceled out. Take the following function for example:
>
> (3*x^2 + 13*x + 4)/(x^3 + 4*x^2 + x + 4) as x-->(-4)
>
> If I'd just plug in -4 into this equation I would get 0/0,
> which I think is called indeterminate form.So I first have
> to cancel common zeros out.
>
> But the thing is, that after I do that it is no longer the
> same function and so the limit also shouldn't be the same.
Good obvervation, however there is a property of limits that says if two
functions agree at all but one point, their limits are the same. IOW, yes,
you are finding the limit of a different function but can conclude this also
to be the limit of the original function.
> Afterall limits for both polynomials(before I cancel zeros
> out) in numerator and denominator are zero,and I
> understand you can't divide zero by zero,but still the
> fact remains that original function is not(or by my
> reasoning shouldn't be) the same as rational function we
> get after we cancel out common zeros.
> I know some will argue that fraction doesn't change if we multiply by same
> number both numerator and denominator ...
Simple example: Let f(x) = x / x^2.
lim x->0 [x / x^2] by direct substitution is the indeterminate form 0/0,
but remember direct substitution is only valid when the function is
*continuous* at the point in question. This function is clearly not
continuous at x=0.
By property of limits, we can cancel the common factor. Granted, we get a
*different* function g(x)=x, but this function g is *continuous* so we can
evaluate its limit by direct substitution:
lim x->0 [x] = g(0) = 0
...and finally, by property of limits we can conclude that lim x->0 f(x)
must also be equal to 0. I believe this property can be extended to cover
any finite number of removable discontinuities between two functions, i.e.
more than one factor cancels between num. and denom.
--
Darrell
Darrell
"Darrell" <dr6...@comcastsnip.net> wrote in message
news:AfGdnR7BlLbzhTje...@comcast.com...
>
> Simple example: Let f(x) = x / x^2.
>
> lim x->0 [x / x^2] by direct substitution is the indeterminate form 0/0,
> but remember direct substitution is only valid when the function is
> *continuous* at the point in question. This function is clearly not
> continuous at x=0.
>
> By property of limits, we can cancel the common factor. Granted, we get a
> *different* function g(x)=x, but this function g is *continuous* so we can
> evaluate its limit by direct substitution:
>
> lim x->0 [x] = g(0) = 0
Oops, that would be true if f(x) = (x^2)/x, but originally I typed
f(x)=x/(x^2). After reducing we have g(x)=1/x and lim x->0 g(x) does not
exist. So let f(x)=(x^2)/x instead, which is what I intended to type in the
first place. Sorry for any confusion.
--
Darrell
boris
bye
boris
>Hey, these are neat! Try rationalizing both the
>numerator and the denominator. For example, with
>the first one, multiply both the numerator and
>denominator by [sqrt(3x+1) + 2][sqrt(x) + 1].
>
>Dave L. Renfro
got it. What bothers me about this is it's all hit and
miss, meaning if you don't find something to cancel out
after factoring it then go back and try multiplying both
the numerator and denominator with something else
Are there any tips for solving these kind of problems or
is that all there is to it?
boris
bye
Rob Johnson
Your message has neither a "References" nor an "In-Reply-To" header.
So without a quotation, no one knows to whom you are talking. The
important part is that you understand limits better and everyone who
has helped you is probably glad.
Dave L. Renfro
>> I almost missed this post. Thank you for your help
Rob Johnson wrote:
> Your message has neither a "References" nor an
> "In-Reply-To" header. So without a quotation,
> no one knows to whom you are talking. The
> important part is that you understand limits
> better and everyone who has helped you is
> probably glad.
What's worse is that if you actually look at the
threading relationship, the post in question is
number 7 in google's sorting of the messages
(as I write this), which means that his post
is a reply to his own post, which means he's
thanking himself for helping him ... ??
Limit of a function-can't solve few problems
1 boris Dec 16
2 Brian M. Scott Dec 16
3 boris Dec 17
4 Dave L. Renfro Dec 17
5 Paul Sperry Dec 17
6 boris Dec 18
7 boris Dec 18
8 Rob Johnson Dec 19
9 Rob Johnson Dec 17
10 boris Dec 17
11 Dave L. Renfro Dec 17
12 Darrell Dec 18
13 Darrell Dec 18
14 boris Dec 18
Dave L. Renfro
boris
>threading relationship, the post in question is
>number 7 in google's sorting of the messages
>(as I write this), which means that his post
>is a reply to his own post, which means he's
>thanking himself for helping him ... ??
>
>http://tinyurl.com/73ls2
>Dave L. Renfro
Well who would have thought Dave, but boris actually doesn't post at that linky you gave. Nah, he posts here
http://mathforum.org/kb/message.jspa?messageID=4149380&tstart=0
and here Rob Johnson's post is displayed as a single thread separated from one I started.
Rob Johnson
Ah, you're welcome.
In any case, I know that there is some way for you to signify the
thread to which you are replying even using MathForum (since there is
a "References" header to your last message). I read news using either
Netscape's newsreader or Google and without a "References" header,
they get confused as to where in the thread, and sometimes to which
thread, a message belongs. These are just courtesies so that readers
know to what your post refers.