Look at the function f(x) = x^(1/2)*[1 - ln(1+ 1/x)^x], and write it as:
f(x) = x^(1/2)*[1 - xln(1 + 1/x)] = [1 - xln(1 + 1/x)]/(1/x^(1/2))
Now apply L'Hospital:
[-ln(1 + 1/x) + 1/(1+x)]/[-(1/2)/x^(3/2)] --> 0...when x --> oo
Regards
Tonio
2007-11-09 01:30:14
·
answer #1
·
answered by Bertrando 4
·
0⤊
0⤋
Remember: the limit of a product is the product of the limits, if both limits exist and are finite. If you are taking the limit as x -> 0, then lim(n^(1/2)) = 0. If you are taking the limit as n -> inf, you have the indeterminate form 0*inf. Write the function in the form [1 - ln((1 + 1/n)^n]/[1/n^(1/2)] and use L'Hospital's Rule.
2007-11-09 01:22:34
·
answer #2
·
answered by Tony 7
·
0⤊
0⤋
it incredibly is no longer any extra reported that (f_n) converges on E, so we could continuously coach this incredibly happens. yet it fairly is easy, by way of certainty, thinking D is dense in E, then E is the closure of D (with understand to E), so as that each and every x ? E is in D or is a decrease returned portion of D. If x is in D, then (f_n(x)) converges by using assumption; in any different case, thinking x is a decrease returned portion of D and (f_n)) converges on D, then, in accordance to a classic theorem, (f_n(x)) converges to 3 incredibly sort. as a consequence, we are able to boost the portion of f to the completed E, which skill that (f_n)) converges on E to the prolonged f. as properly, by way of certainty the purposes f_n are non-furnish up on E - so on D - and (f_n) converges uniformly to f on D, then, in accordance to a various classic theorem, f is non-furnish up on D. yet, for now, we don't understand if f is non-furnish up on the completed E. to coach the convergence is uniform on E, enable ? > 0 and, based on the uniform convergence on D, %. N such that |f_n(x) - f(x)| < ?/3 for each x ? D on each and every occasion n > N. enable n > N. thinking D is dense in E and f_n and f are non-furnish up on D, we are able to come across y ? D such that |f_n(y) - f_n(x)| < ?/3 and |f(y) - f(x)| < ?/3. y relies upon on ?, x and n, yet this does not remember by way of certainty, for each ?, x and n, there is such y. Then, |f_n(x) - f(x)| ? |f_n(x) - f_n(y)| + |f_n(y) - f(y)| + |f(y) - f(x)| < ?/3 + ?/3 + ?/3 = ?. thinking this holds for each x ? E (N relies upon basically on ?) the convergence f_n ? f is uniform on E. and since the purposes f_n are non-furnish up on E and (f_n) converges uniformly to f on E, f is non-furnish up on E.
2016-11-10 22:02:25
·
answer #3
·
answered by Anonymous
·
0⤊
0⤋