Help with some questions?:)

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- Jan 4th 2007, 01:30 PMsbsitelimits?
Help with some questions?:)

- Jan 4th 2007, 03:21 PMThePerfectHacker
Problem #2

Expand as a binomial,

Expand as a binomial,

Expand as a binomial,

Thus, we have,

Take the Cauchy product,

Only the first coefficients are of interest,

Thus,

- Jan 4th 2007, 04:05 PMgalactus
For #3, we can use the ol' Hospital rule.

L'Hopital, derivative of num. and den.:

=

For #6:

There are some identities we can employ:

Believe it or not:

Now, we have a nice, little, itty-bitty, easy limit.

- Jan 5th 2007, 04:28 AMtopsquark
For number 1 we can try L'Hopital's rule since the numerator and denominator both go to 0:

-Dan - Jan 5th 2007, 04:44 AMtopsquark
For 5 there are other ways to do this (like L'Hopital's rule), but you can use one of my favorite methods. Since we are looking at x close to 0, use a Taylor's expansion of the numerator and denominator around x = 0:

So:

Now, the denominator goes to 0 while the numerator goes to . Presuming that the limit is infinite (ie doesn't exist.)

If then

-Dan - Jan 5th 2007, 04:48 AMtopsquark
#4. I don't have time for full rigor here, but since for all x > 1 I would say this limit is infinite. To get a proof, note that the limit is of the form . You can convert this to either or and then use L'Hopital's rule.

-Dan - Jan 5th 2007, 05:22 AMSoroban
Hello, sbsite!

Here's #2 . . . without the Binomial Expansion.

Quote:

Divide top and bottom by

. .

. .

Then: .

- Jan 5th 2007, 06:01 AMSoroban
Hello again, sbsite!

A couple more . . . without using L'Hopital.

Quote:

We have: .

Multiply top and bottom by

. .

Therefore: .

Quote:

Divide top and bottom by

Multiply the first fraction by , the second fraction by

. .

Then: .

- Jan 5th 2007, 06:33 AMtopsquark
#4 Addendum. I was thinking to factor a from the first term and factor, and use the fact that the limit now takes the form of . You can convert this to the form and use L'Hopital's rule on it, but we now get an indeterminate form , and it doesn't look another application of L'Hopital's rule is going to help.

Perhaps this will work and I'm simply not being persistant enough, but I'm assuming another method (which I can't think of at the moment) will produce easier results.

(Note: Numerical estimates suggest that the limit is .)

-Dan - Jan 5th 2007, 06:52 AMTD!
For #7:

Substitute x = y + 1, then you have to take the limit for y to 0.

Note that tan(pi/2+a) = -cot(a), so we can simplify like this:

Now, arround 0, Taylor says that ; so:

- Jan 5th 2007, 07:16 AMTD!
For #4:

To avoid l'Hopital with all these roots, multiply numerator and denominator with the complement:

Simplify in the numerator ((a-b)(a+b) = a²-b²) and divide numerator and denominator by sqrt(x):

Keep shifting the x in the square roots (sqrt(a)/x = sqrt(a/x²) with x positive):

Now, all the fractions 1/x and the one 1/x^(3/2) go to zero for x to infinity, so:

- Jan 5th 2007, 10:47 AMTD!
Addendum for #7.

If you can't (or rather not) use Taylor, the last steps can be replaced by using the standard limit of tan(x)/x for x to 0.

That limit is of course, just as the one for sin(x)/x with x to 0, 1. In that case, write:

- Jan 5th 2007, 02:45 PMgalactus
Here's a method I used for the nested radical. It's not as eloquent as TD's, but what the heck:

Let

With some algbraic gymnastics, we get:

and

And we have:

............................................ - Jan 5th 2007, 03:52 PMSoroban
Hello, sbsite!

I*think*I have #4 . . .

Quote:

Let

We have: .

Multiply top and bottom by

. .

Divide top and bottom by

. . . . .

We have: .

- Jan 6th 2007, 01:21 AMTD!
Soroban, we did exactly the same, only you named sqrt(x) = u. The steps are the same ;)