orivd
幼苗
共回答了30个问题采纳率:96.7% 举报
limn→∞ [ln(√n^2+1)-ln(n+1)]/(1/n)
=limn→∞ [n/(n^2+1)-1/(n+1)]/(-1/n^2)
=limn→∞ [n^2(1-n)]/[(n^2+1)(n+1)]
=limn→∞ (1/n-1)/[(1+1/n^2)(1+1/n)
=(0-1)/(1+0)(1+0)
=-1,
——》原式=limn→∞e^ln[√(n^2+1)/(n+1)]^n
=e^limn→∞ [ln(√n^2+1)-ln(n+1)]/(1/n)
=e^(-1).
1年前
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