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f(x) =arctan[ 1/(x-1)] + sinx/[x^2.(π-x)]
lim(x->1-) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] }
=-π/2 + sin1/(π-1)
lim(x->1+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] }
=π/2 + sin1/(π-1)
x=1 : 跳跃间断点
lim(x->0+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] } ->+∞
x=0 : 无穷间断点
lim(x->π+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] } ->+∞
x=π : 无穷间断点
lim(x->1-) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] }
=-π/2 + sin1/(π-1)
lim(x->1+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] }
=π/2 + sin1/(π-1)
x=1 : 跳跃间断点
lim(x->0+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] } ->+∞
x=0 : 无穷间断点
lim(x->π+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] } ->+∞
x=π : 无穷间断点
更多追问追答
追问
x趋向于0时,=∞是怎么想出来的?
追答
lim(x->0+) { arctan[ 1/(x-1)] + sinx/[x^2.(π-x)] }
=lim(x->0+) { arctan[ 1/(x-1)] + 1/[x.(π-x)] } ->+∞
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