帮忙看下这个式子极限怎么求当x趋于无穷时候[(x^2-1)/(x^2+1)]^x^2的极限 利用罗比达法则求 麻烦写详细
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令y=[(x^2-1)/(x^2+1)]^x^2

lny=x^2ln[(x^2-1)/(x^2+1)]

令x^2=1/t

则x→∞即t→0

lim(x→∞)x^2ln[(x^2-1)/(x^2+1)]

=lim(x→∞){ln[(t-1)/(t+1)]}/t

=lim(x→∞)[(t+1)/(t-1)]*[2/(t+1)^2]

=lim(x→∞)2/(t^2-1)

=-2

所以原极限=1/e^2

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