3√(x3-5x2+1))-x=3√(1-5/x+1/x3))-1]/(1/x)用诺必达法则,这是一个0/0型的极限,分子分母求导,有lim(x->∞)((3√(x3-5x2+1))-x) =lim(x->∞)[3√(1-5/x+1/x3))-1]/(1/x)]=lim(x->∞)[[(-3/x^4+5/x^2) / 3[3√(1-5/x+1/x3))]2]/(-1/x^2)]=lim(x->∞)[(3/x^2-5) / 3[3√(1-5/x+1/x3))]2]=-5/3construct a table to evaluate the values of the limit.lim(x-->0?) (1+5x)^(1/5x)values of x :-1 -0.1 -0.01 -0.001 -0.0001 0values of (1+5x)^(1/5x):-0.7589 4.0000 2.7895 2.7251 2.7190 undefinedvalues of x :0.0001 0.001 0.01 0.1 1values of (1+5x)^(1/5x):2.7176 2.7115 2.6533 2.2500 1.4310since e ≈ 2.718...therefore the limit of (1+5x)^(1/5x) as x approaches zero from the right,is e.where e is the base of the natural logarithm.