Question

急求高数极限解法 lim(x→0)ln(1+2x^2)/xsinx 好像不能用罗比达法则,

Answer 1

∵原式=lim(x->0){[ln(1+2x²)/x²]*(x/sinx)}
=lim(x->0)[ln(1+2x²)/x²]*lim(x->0)(x/sinx)
又lim(x->0)[ln(1+2x²)/x²]=lim(x->0)[(4x/(1+2x²))/(2x)] (0/0型,应用罗比达法则)
=lim(x->0)[2/(1+2x²)]
=2
lim(x->0)(x/sinx)=lim(x->0)[1/(sinx/x)]
=1/[lim(x->0)(sinx/x)]
=1/1 (应用重要极限lim(x->0)(sinx/x)=1)
=1
∴原式=2*1=2.