若limx-＞1x2+ax+b除以sin(x2-1)=3则a=?b=?请问sinx2-1怎么有理化呢?

Question

若limx-＞1 x2+ax+b除以sin(x2-1)=3则a=?b=?
请问sinx2-1怎么有理化呢?

Answer 1

x→1时（x^2+ax+b)/sin(x^2-1)
→(x^2+ax+b)/[(x+1)(x-1)]→3,
∴x-1|x^2+ax+b,
由余数定理,1+a+b=0,b=-1-a,
x^2+ax-1-a=(x-1)(x+1+a),
原式变为(x+1+a)/(x+1)→(2+a)/2=3,
∴a=4,b=-5.
sin(x^2-1)用等价无穷小量(x^2-1)代替,不是有理化.