Re: more help needed w/proving limit of a sequence
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Sun, 14 Sep 2008 02:36:14 EDT
On Sep 13, 2008 11:44 PM CT, TheGist wrote:
Thanks for all the help so far. I think I have a good
solution but do
not have the confidence yet to think I am 100% right.
Here is the question and my solution...
Show lim sqrt(3+(n/2))=sqrt(3)
Stop right there! You're evaluating the limit as n
approaches what?
The only way the above limit would be valid would be if
you were considering the limit of the function f(n) =
sqrt(3 + n/2) as n --> 0...
...but you're looking at the limit of sequences, so I'm
assuming you've made a typo and your question should be:
Show lim_{n --> oo} sqrt(3 + 2/n) = sqrt(3).
Solution:
Solving for an N to show |X_n - L|< epsilon...
Multiplying by the conjugate
(sqrt(3+(n/2)) - sqrt(3))
* (sqrt(3+(n/2)) + sqrt(3))
-----------------------------
3 + (2/n) - 3
No, according to what you've written, the answer should
be 3 + (n/2) - 3 = n/2...
...but you've probably switched the n and 2 around.
Now, also dividing by the conjugate we
end up with
(2/n)
-------------------------
(sqrt(3+(n/2)) + sqrt(3))
No, this should either be
(n / 2) / [sqrt(3 + n/2) + sqrt(3)]
...or...
(2 / n) / [sqrt(3 + 2/n) + sqrt(3)].
I'm assuming it's the latter.
We note that the numerator has an upper bound of n
since (2/n) < n for all n Natural greater than 2
While it is true that 2/n < n for n > 2 (since 2/x < x
for all real numbers x > sqrt(2)), you shouldn't say
that n is the "upper bound" of the numerator -- upper
and lower bounds of sequences have well established
definions.
and the denominator has a lower bound of sqrt(3)
since (sqrt(3+(n/2)) + sqrt(3)) > sqrt(3) for all n
Natural
...again, you should probably stay away from that
terminology (unless you use it) and state the inequality.
So, now for the "epsilontics"...
|X_n -L | < n/sqrt(3) < epsilon
So if we let N be a natural number such that
N> 1/(sqrt(3)*epsilon) then for all n > N the stated
limit holds.
Is this solution acceptable?
If not, could you please tell me why?
I suppose that all depends on your definition of
"acceptable" : - P The statement of your problem and
your solution both had various typos; moreover, as a
whole, the proof doesn't flow very well.
In other words, you may want to polish it up before you
hand it in for homework.
I am very uncertain about my estimated bounds...is
this the correct approach to take? Obviously, the
choices I made for the bounds is very loose but isn't
it acceptable to not necessarily have tight bounds
when doing these sorts of proofs?
I'm not sure what exactly to tell you here...
There may be eaiser ways to write the same proof using
alternate or fewer inequalities, but the bottom line is
that if an inequality is valid, you can use it in your
proof.
Regards,
Kyle Czarnecki
P.S. Whenever you've completed an epsilon-delta proof,
you can "check" with a few numerical examples.
For instance, consider the following exercise/proof.
EXERCISE. Show that lim_{n -> oo} sqrt(3 + 2/n) = sqrt(3)
PROOF. Given e > 0, choose a natural number N such that
N > 2 / [sqrt(3) * e]. By definition
|x_n - L| = |sqrt(3 + 2/n) - sqrt(3)|
= sqrt(3 + 2/n) - sqrt(3)
= (2 / n) / [sqrt(3 + 2/n) + sqrt(3)]
< (2 / n) / sqrt(3)
= 2 / [n * sqrt(3)].
Moreover, since N > 2 / [sqrt(3) * e], it follows that
|x_n - L| < 2 / [n * sqrt(3)]
= (1 / n) * [2 / sqrt(3)]
< (1 / n) * (N * e)
...and thus |x_n - L| < e whenever n > N. []
Alright, now let's "check" the proof via a numerical
example.
Suppose I choose epsilon (e) such that
e = 1/37 = 0.027027...
...then, according to the proof, whenever
n > N > 2 / [sqrt(3) * e] = 2 * 37 / sqrt(3) = 42.72...
...we should have...
|sqrt(3 + 2/n) - sqrt(3)| < e = 0.027027...
Let f(n) = sqrt(3 + 2/n) - sqrt(3), then
f(43) = 0.013375... < e = 1/37,
f(44) = 0.013072... < e = 1/37,
f(45) = 0.012783... < e = 1/37, etc.
...so the proof "works" for this choice of epsilon.
Moreover, note that |f(n)| < e implies that
-e < f(n) < e
-e < sqrt(3 + 2/n) - sqrt(3) < e
sqrt(3) - e < sqrt(3 + 2/n) < sqrt(3) + e
...or, in other words, sqrt(3 + 2/n) lies in the interval
(sqrt(3) - e, sqrt(3) + e).
In this specific case where e = 1/37, this means that
sqrt(3 + 2/n) will lie in the interval (sqrt(3) - 1/37,
sqrt(3) + 1/37) whenever n > 42.27...
.
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- more help needed w/proving limit of a sequence
- From: TheGist
- more help needed w/proving limit of a sequence
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