I would like to reply in english, as slow in typing chinese.
all of these will be the application of Lebesgue’s Dominated Convergence Theorem.
1)
Int[ (1+x/n)^n * e^{-2x}, 0, n ] = Int[ f_n(x), 0, \infty]
where f_n(x)= Character(x, [0,n]) * (1+x/n)^n * e^{-2x} , which is pointwise convergent to e^{-x}. Now show that f_n(x) <= e^{-0.5 x}. Or, you only need to show that when t=x/n, 1+t < e^{1.5t} for all t>0.
Then since F(x)=e^{-0.5 x} is integrable on (0, infty), so the limit can be moved into the integral and the answer is int( e^{-x} , 0, \infty)=1.
2)let t=x/n. then it becomes int[sin(t)*n/ (1+t)^n; 0, infty].
int[sin(t)*n/ (1+t)^n] = (integral by parts) = -n* sint /(1+t)^{n-1} /(n-1)
+ n /(n-1) Int[ cos(t)/(1+t)^{n-1}] The first part, the function -n* sint /(1+t)^{n-1} /(n-1) evaluates to 0 at 0 and infty. The second integral Int[ cos(t)/(1+t)^{n-1}; 0, infty] is absolutely less than Int[ 1/(1+t)^{n-1}; 0, infty]= 1/(n-2).
So combine together, and let n go to infty, the answer is 0.
3). Notice that the function sin(t)/t is continuous over R and is bounded by 1. therefore, the integral is directly bounded by 1/(1+x^2). Therefore, the limit can be taken inside the integral, and it becomes Int[ 1/(1+x^2); 0, infty] = pi/2.
4) it has anti-derivative arctan(nx). arctan(nx) at infty is pi/2. arctan(nx) at a depends on a. If a is positive, then the value is pi/2. If a=0, then the value is 0; if a is negative, then the value is -pi/2.
So by subtraction, you will see that there are three answers depending the sign of a: 0, pi/2, and pi.
答案: (1): 1. (2): 0. (3): pi/2. (4): o or pi/2 or pi.
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