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极限lim(x->3)根号下[(x-3)/(x^2-9)]怎么求?
题目详情
极限lim(x->3) 根号下[(x-3)/(x^2-9)]怎么求?
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答案和解析
分母因式分解
x^2-9=x^2-3^2=(x-3)(x+3)
原式=(x-3)/[(x-3)(x+3)]=1/(x+3)
所以
lim(x->3) 根号下[(x-3)/(x^2-9)]
=lim(x->3) 根号下(1/(x+3))
=根号(1/(3+3))
=根号(1/6)
=(根号6)/6
x^2-9=x^2-3^2=(x-3)(x+3)
原式=(x-3)/[(x-3)(x+3)]=1/(x+3)
所以
lim(x->3) 根号下[(x-3)/(x^2-9)]
=lim(x->3) 根号下(1/(x+3))
=根号(1/(3+3))
=根号(1/6)
=(根号6)/6
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