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,
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.
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
#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
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: