Expand according to Taylor series

cos(x^2)= 1-x^4/2+x^8/24+o(x^6)

And sin (x 2) = x 2-x 6/6+o (x 6).

Bring to the limit

=lim[x^2( 1-x^4/2+x^8/24+o(x^6))-(x^2-x^6/6+o(x^6)]/(x^6)

=- 1/3

2

answer

∫cosx/(2+sinx)^2 dx=∫d(2+sinx)/(2+sinx)^2=- 1/(2+sinx)

Because sinx, when x tends to 0, is oscillating, not a certain value in the convergence domain, so it cannot be integrated.

b)

Although x/(1-x 2) tends to infinity when x=- 1 and1,the integral seems not to converge, but actually x/(1-x 2).

It is a odd function at (-1, 1), so the area enclosed by it and the X axis is negative at (-1, 0) and positive at (0, 1).

The sum is 0. So the integral is 0. Above:

three

a

a(k)=ln( 1+k)x^k/e^k

Let lim | a (k+1)/a (k) | = | x/e | <1

| x | & lte

The convergence radius r = e is obtained.

b)

Because lim[ln( 1+k)]= infinity,

When x=e and -e, the original series does not converge, so the convergence domain is (-e, e).