lim/x→3.14/2^+ ln(x-3.14/2)/tanx

lim/x→3.14/2^+ ln(x-3.14/2)/tanx
lim/x→3.14/2^+ ln(x-3.14/2)/tanx

lim/x→3.14/2^+ ln(x-3.14/2)/tanx
lim(x->π/2+)[ln(x-π/2)/tanx]=lim(x->π/2+){[1/(x-π/2)]/sec²x} (∞/∞型极限,应用罗比达法则)
=lim(x->π/2+)[cos²x/(x-π/2)] (0/0型极限,应用罗比达法则)
=lim(x->π/2+)(-2sinx*cosx)
=lim(x->π/2+)[-sin(2x)]
=-sinπ
=0.

lim/x→π/2+/[ln(x-π/2)/tanx]属于lim(∞/∞)模型
∴由洛必达法则，得 原式=lim/x→π/2+/ {[1/(x-π/2)]/[1+(tanx)^2]}
=lim/x→π/2+/ {1/(x-π/2)[1+(tanx)^2]}
=lim/x→π/2+/ [1/(x-π/2)]
=+∞它等于0呀原式=lim/x→π/2+/ [1/(x-...

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lim/x→π/2+/[ln(x-π/2)/tanx]属于lim(∞/∞)模型
∴由洛必达法则，得 原式=lim/x→π/2+/ {[1/(x-π/2)]/[1+(tanx)^2]}
=lim/x→π/2+/ {1/(x-π/2)[1+(tanx)^2]}
=lim/x→π/2+/ [1/(x-π/2)]
=+∞

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