lim(x->无穷)[ln(1+1/x)]/[arc cotx]用洛必达法则求

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lim(x->无穷)[ln(1+1/x)]/[arccotx]用洛必达法则求lim(x->无穷)[ln(1+1/x)]/[arccotx]用洛必达法则求lim(x->无穷)[ln(1+1/x)]/[a

lim(x->无穷)[ln(1+1/x)]/[arc cotx]用洛必达法则求
lim(x->无穷)[ln(1+1/x)]/[arc cotx]用洛必达法则求

lim(x->无穷)[ln(1+1/x)]/[arc cotx]用洛必达法则求
lim(x->无穷)[ln(1+1/x)]/[arc cotx]
分子分母同时求导得lim(x->无穷)[ln(1+1/x)]/[arc cotx]
=lim(x->无穷)[[-1/(x^2+x)]/[-1/(1+x^2)] =lim(x->无穷)(1+x^2)/(x^2+x)=1