lim(n→∞) {[4/5-6/7+4/(5^2)-6/(7^2)+...+4/(5^n)-6/(7^n)]/[5/6-4/5+5/(6^2)-4/(5^2)+...+5/(6^n)-4/(5^n)]}=( )A.-1 B.0 C.1 D.6/7

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lim(n→∞){[4/5-6/7+4/(5^2)-6/(7^2)+...+4/(5^n)-6/(7^n)]/[5/6-4/5+5/(6^2)-4/(5^2)+...+5/(6^n)-4/(5^n)]

lim(n→∞) {[4/5-6/7+4/(5^2)-6/(7^2)+...+4/(5^n)-6/(7^n)]/[5/6-4/5+5/(6^2)-4/(5^2)+...+5/(6^n)-4/(5^n)]}=( )A.-1 B.0 C.1 D.6/7
lim(n→∞) {[4/5-6/7+4/(5^2)-6/(7^2)+...+4/(5^n)-6/(7^n)]/[5/6-4/5+5/(6^2)-4/(5^2)+...+5/(6^n)-4/(5^n)]}=( )
A.-1 B.0 C.1 D.6/7

lim(n→∞) {[4/5-6/7+4/(5^2)-6/(7^2)+...+4/(5^n)-6/(7^n)]/[5/6-4/5+5/(6^2)-4/(5^2)+...+5/(6^n)-4/(5^n)]}=( )A.-1 B.0 C.1 D.6/7
先算4/5+4/(5^2)+...+4/(5^n),通分得{4*[1+5+25+...5^(n-1)]}/5^n
约分得到[(5^n)-1]/5^n.
同理其它的都可以这么算,最后的lim(n→∞){[4/5-6/7+4/(5^2)-6/(7^2)+...+4/(5^n)-6/(7^n)]/[5/6-4/5+5/(6^2)-4/(5^2)+...+5/(6^n)-4/(5^n)]}等于lim(n→∞){{[(5^n)-1]/5^n}-{[(7^n)-1]/7^n}}/{{[(6^n)-1]/6^n}-{[(7^n)-1]/7^n}}由洛必达定理得结果是-1!选A
其实选择题不用这么算,因为观察看分子是一个负数,分母是一个正数,最后结果一定是负数,选项只有A是负数,所以只能选A!

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