Question

1. Use the Taylor series to calculate the limit.

Problem: limit as x approaches 0 is equal to (1-cos(x))/(1+x-e^x).

I did the problem out but I need help in seeing if its correct.

limit as x approaches 0 is equal to (1-cos(x))/(1+x-e^x)= (1-(1-x^2/2+x^4/4!-...))/(1+x-(1+x+x^2/2+x^3/3!+x^4/4!+...)=(x^2-x^4/4!+...)/(-x^2/2-x^3/3!-x^4/4!-...)=-11/17

Answer 1

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:

1-cos(x) = 1/2 x^2 + O(x^4)

1 +x-exp(x) = -1/2 x^2 + O(x^3)

[1-cos(x)]/[1+x-exp(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.