Question

高数极限问题求解惑

请问高手，画横线的这两步是怎么得到的？求解惑

Answer 1

lim{x->1} [x^2+mx+n]/[sin(x^2-1)] = 3
因为分母趋于零，极限值是常数，意味着分子必须和分母同阶无穷小，所以有 m+n+1 = 0
lim{x->1} [x^2+mx+n]/[sin(x^2-1)]
= lim{x->1} [x^2+mx+n]/[(x^2-1)] (等价无穷小代换）
= lim{x->1} [x^2+mx+n]/[(x+1)(x-1)]
= lim{x->1} 1/(x+1) * lim{x->1} [x^2+mx+n]/(x-1)
= (1/2) lim{x->1} [x^2+mx+n]/(x-1)
= (1/2) lim{x->1} [2x+m]
= (1/2)(2+m)
从 (1/2)(2+m) = 3，解得：m = 4
再从m+n+1 = 0可得：n = -5

Answer 2

分母趋近于0，极限存在，分子一定趋近于0
先分解因式，再把得到常数的极限提取出来