
求lim(x->3)|[√(x+1)-2]/3-√(2x+3)的极限
2个回答
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解:
lim [√(x+1)-2]/[3-√(2x+3)]
x→3
=lim [√(x+1)-2][√(x+1)+2][3+√(2x+3)] / [3-√(2x+3)][3+√(2x+3)][√(x+1)+2]
x→3
=lim [(x+1)-4][3+√(2x+3)] / [9-(2x+3)][√(x+1)+2]
x→3
=lim (x-3)[3+√(2x+3)] / (-2)(x-3)[√(x+1)+2]
x→3
=lim [3+√(2x+3)] / (-2)[√(x+1)+2]
x→3
=[3+√(2·3+3)] / (-2)[√(3+1)+2]
=-¾
lim [√(x+1)-2]/[3-√(2x+3)]
x→3
=lim [√(x+1)-2][√(x+1)+2][3+√(2x+3)] / [3-√(2x+3)][3+√(2x+3)][√(x+1)+2]
x→3
=lim [(x+1)-4][3+√(2x+3)] / [9-(2x+3)][√(x+1)+2]
x→3
=lim (x-3)[3+√(2x+3)] / (-2)(x-3)[√(x+1)+2]
x→3
=lim [3+√(2x+3)] / (-2)[√(x+1)+2]
x→3
=[3+√(2·3+3)] / (-2)[√(3+1)+2]
=-¾
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