Posted by Joshua on Tuesday, April 28, 2009 at 11:57pm.
1. Use the Taylor series to calculate the limit.
Problem: limit as x approaches 0 is equal to (1cos(x))/(1+xe^x).
I did the problem out but I need help in seeing if its correct.
limit as x approaches 0 is equal to (1cos(x))/(1+xe^x)= (1(1x^2/2+x^4/4!...))/(1+x(1+x+x^2/2+x^3/3!+x^4/4!+...)=(x^2x^4/4!+...)/(x^2/2x^3/3!x^4/4!...)=11/17

Calculus  Count Iblis, Wednesday, April 29, 2009 at 8:50am
It is wrong. You also need to use the "Big O" notation. One denotes by
O(x^n) a term that for x to zero behaves as a constant times x^n. More rigorously, we say that some term T(x) is O(x^n) if and only if there exists a p and C such that
T(x) <= C x^n for all x < p
THen we can proceed as follows. We have:
1cos(x) = 1/2 x^2 + O(x^4)
1 +xexp(x) = 1/2 x^2 + O(x^3)
[1cos(x)]/[1+xexp(x)] =
[1+O(x^2)]/[1+O(x)] =
[1+O(x^2)][1+O(x)] =
1 + O(x)
Limit of x to zero is then 1, because the O(x) tends to zero as it is less than x for x sufficiently close to zero.
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