求极限请给出过程 lim x-> 1+ ln(sinh(x-1))/ln(x^2-1)
1个回答
用罗比达法则
=lim[cosh(x-1)/sinh(x-1)]/[2x/(x^2-1)]
=(1/2)lim(x^2-1)/sinh(x-1)
=limx/cosh(x-1)
=1
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