高数极限题目求详解lim(x→0）[√（1+xtanx)]-1/[e^(x^2)]-1 ,答案是1/2有答案,但是求详细

题目是:(√(1+x·tan(x))-1)/(e^(x^2)-1)
首先,x → 0时,y = x·tan(x) → 0,因此(√(1+y)-1)/y → 1/2 (基本极限),
即f(x) = (√(1+x·tan(x))-1)/(x·tan(x)) → 1/2.
其次,x → 0时,z = x^2 → 0,因此(e^z-1)/z → 1 (基本极限),
即g(x) = (e^(x^2)-1)/x^2 → 1.
最后,x → 0时,h(x) = tan(x)/x = 1/cos(x)·sin(x)/x → 1.
综合得:(√(1+x·tan(x))-1)/(e^(x^2)-1) = f(x)h(x)/g(x) → 1/2·1/1 = 1/2.