| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 921.1 | What's in a limit anyway? | POOL::HALLYB | The smart money was on Goliath | Wed Aug 24 1988 17:28 | 21 |
| 921.2 | I Think that... | LANDO::ICOHEN | | Wed Aug 24 1988 17:36 | 7 |
| I'm not a math guru, but I think the answer is along these lines:
When you let x=0 you would get 0/0. In that case to find the limit
use L'Hopital's rule which says take the derivative of the numerator
and the denominator and then let x->0. This gives 1/(2*SQRT(6)).
|
| 921.3 | looked up L'Hopital, but... | VLSBOS::HARTEL | | Wed Aug 24 1988 17:58 | 9 |
|
This is from a calc 1 class; L'Hopital's Rule isn't touched on until
calc 3. I looked at it though, and while it did make sense, it
didn't help me correctly solve the problem. I understand what you
said about the limit not actually being 0, but that still doesn't
change my answer. Any other suggestions?
Cathy
|
| 921.4 | another way | LISP::DERAMO | Daniel V. {AITG,LISP,ZFC}:: D'Eramo | Wed Aug 24 1988 22:52 | 26 |
| If you expand sqrt(x + 6) using the binomial theorem
you get [that is, one way of doing it gets]
1/2 1/2 -1/2 -3/2 2
(6 + x) = 6 + (1/2)6 x + (1/2)(-1/2)6 x + ...
m
where the terms in ... involve x for m >= 3. This
infinite sum converges for small enough values of x [and
converges to the expected value].
1/2
If you now subtract 6 and divide by x you get
-1/2 -3/2
f(x) = (1/2)6 - (1/4)6 x + ...
m
where here the terms in ... involve x for m >= 2.
Now as x -> 0, this last expression shows that the function
approaches the value given by others,
-1/2
(1/2)6
Dan
|
| 921.5 | Thanks | VLSBOS::HARTEL | | Thu Aug 25 1988 03:20 | 6 |
|
Now I understand. Guess it was easy after all!
Cathy
|
| 921.6 | | BEING::POSTPISCHIL | Always mount a scratch monkey. | Thu Aug 25 1988 12:21 | 20 |
| Another way to do it is to return to the definition of limit. Suppose
that [sqrt(x+6)-sqrt(6)]/x = L + e for x's near zero and small e's
(positive or negative). Simply solve for x to get:
[1 - 2 sqrt(6) (L+e)] / (L+e)^2 = x.
(To get there from the previous equation, multiply both sides by x,
square, cancel the 6's on each side, divide by x, and you now have a
linear equation in x.)
Since (L+e)^2 is positive, to get the above expression near zero, you
have to make 1 - 2 sqrt(6) (L+e) close to zero, and it is pretty
obvious what value of L does that.
This can also be converted directly into a proof, since any desired
maximum error, e, gives you a range of x's near zero that produce that
error or less.
-- edp
|
| 921.7 | Rationalize the numerator | PNEUMA::ROOS | | Thu Aug 25 1988 12:39 | 16 |
|
Another way to obtain the solution is to rationalize the numerator.
In this case that requires multiplying the numerator and denominator
by sqrt(x + 6) + sqrt(x). The result of this would be:
Limit x
x->0 ---------------------
x(sqrt(x+6) + sqrt(x))
which simplifies to Limit 1
x->0 ------------------
sqrt(x+6) + sqrt(x)
This limit is 1/(2*sqrt(x)).
|
| 921.8 | | CTCADM::ROTH | If you plant ice you'll harvest wind | Thu Aug 25 1988 15:25 | 7 |
| Geometrically, you have two curves which happen to intersect at the
origin... so the limiting behaviour can be visualized in that neighborhood
as a pair of intersecting tangent lines with (possibly) different slopes.
I find this easier to remember than a specific formula.
- Jim
|
| 921.9 | (a+b)(a-b)=a^2-b^2 | EAGLE1::BEST | R D Best, sys arch, I/O | Thu Aug 25 1988 17:07 | 18 |
| >
> ______ ___
> Here's the problem: Lim V x + 6 - V 6
> x->0 ----------------
> x
>
> ___
> The answer is given as: 1/2 V 6
>
>
Multiply top and bottom by
________ ____
V x + 6 + V 6
and divide numerator and denominator by x.
The resulting expression can be evaluated at x= 0.
|