Question

求解高数关于极限的三道题，要步骤的详细点啊！！有图，在线等！

Answer 1

看看吧

Answer 2

[(2+x)/(2-x)]^(1/x)=[1 + 2x/(2-x)]^(1/x) = [1+2x/(2-x)]^[(2-x)/(2x)*2/(2-x)] = {[1+2x/(2-x)]^[(2-x)/(2x)]}^[2/(2-x)],
lim_{x->0}[2/(2-x)]=1,
lim_{x->0}[2x/(2-x)]=0, lim_{x->0}[1+2x/(2-x)]^[(2-x)/(2x)] = e,
原式=e^1 = e.

lim_{x->0}[3^x-1]=0, 所以, 0=lim_{x->0}ln[1+f(x)/sin(2x)],
1=lim_{x->0}ln[1+f(x)/sin(2x)]/[f(x)/sin(2x)]=lim_{x->0}[f(x)/sin(2x)]/[f(x)/(2x)],
3^x-1=e^[xln(3)] - 1,
1=lim_{x->0}{e^[xln(3)]-1}/[xln(3)]=lim_{x->0}[3^x-1]/[xln(3)],
5=lim_{x->0}ln[1+f(x)/sin(2x)]/[3^x-1] = lim_{x->0}[f(x)/(2x)]/[3^x-1]=lim_{x->0}[f(x)/(2x)]/[xln(3)]
=lim_{x->0}f(x)/[2x^2ln(3)]
10ln(3)=lim_{x->0}f(x)/x^2

lim_{x->无穷}sin(x)/x^2 = 0 = lim_{x->无穷}cos(x)/x^2,
原式=lim_{x->无穷}[1+sin(x)/x^2]/[2-cos(x)/x^2] = [1+0]/[2-0] = 1/2