Please Find:
lim x->2
x-2/√(x²-2)-√(2)
limit
2010-05-11 3:52 pm
回答 (2)
2010-05-11 4:04 pm
✔ 最佳答案
lim x→2 (x - 2) / [√(x^2 - 2) - √2]= lim x→2 (x - 2)[√(x^2 - 2) + √2] / [(x^2 - 2) - (2)]
= lim x→2 (x - 2)[√(x^2 - 2) + √2] / (x^2 - 4)
= lim x→2 (x - 2)[√(x^2 - 2) + √2] / (x - 2)(x + 2)
= lim x→2 [√(x^2 - 2) + √2] / (x + 2)
= [√(2^2 - 2) + √2] / (2 + 2)
= 2√2 / 4
= 1 / √2
參考: Physics king
2010-05-16 7:53 am
Method 1
= lim x->2 (x-2)( √(x²-2)+ √(2) ) / (x²-2) -- (2)
分子分母同時乘 ( √(x²-2)+ √(2) )
= lim x->2 (x-2)( √(x²-2)+ √(2) ) / ( x² -4 )
( a² -b² ) =(a+b)(a-b)
= lim x->2 (x-2)( √(x²-2)+ √(2) ) / ( x+2 )( x-2 )
= lim x->2 ( √(x²-2)+ √(2) ) / ( x+2 ) ----- (*)
=2√(2) /4
sub x=2 into (*)
=√(2) /2
Method2
= lim x->2 (1) / { 1/2 [ ( x²-2 )^-1/2 ] [ 2x ] }
分子分母「各自」微分 using L'Hopital's Rule
= lim x->2 √( x² - 2 ) / x
sub x=2
=√(2) /2
= lim x->2 (x-2)( √(x²-2)+ √(2) ) / (x²-2) -- (2)
分子分母同時乘 ( √(x²-2)+ √(2) )
= lim x->2 (x-2)( √(x²-2)+ √(2) ) / ( x² -4 )
( a² -b² ) =(a+b)(a-b)
= lim x->2 (x-2)( √(x²-2)+ √(2) ) / ( x+2 )( x-2 )
= lim x->2 ( √(x²-2)+ √(2) ) / ( x+2 ) ----- (*)
=2√(2) /4
sub x=2 into (*)
=√(2) /2
Method2
= lim x->2 (1) / { 1/2 [ ( x²-2 )^-1/2 ] [ 2x ] }
分子分母「各自」微分 using L'Hopital's Rule
= lim x->2 √( x² - 2 ) / x
sub x=2
=√(2) /2
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