lim(X-∞)X[√(x2+1)-x]请问这道极限题怎么算啊
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lim(X-∞)X[√(x2+1)-x]
请问这道极限题怎么算啊
答:
因为:[√(x^2+1)-x ]*[√(x^2+1)+x]=x^2+1-x^2=1
所以:
lim(x→∞) x[√(x^2+1)-x]
=lim(x→∞) x/[√(x^2+1)+x]
=lim(x→∞) x/(x+x)
=1/2
中括号内利用平方差公式写成(X²+1-X)/(根号(X²+1)+X),再上下约分X,最后用无穷大代入X
分子分母同时乘以 [√(x2+1)+x],则原式化为x/[√(x2+1)+x],当x->∞时,极限为1/2
lim(x->∞)x[√(x^2+1)-x]
=lim(x->∞)x [√(x^2+1)-x][√(x^2+1)+x]/ [√(x^2+1)+x]
=lim(x->∞) x / [√(x^2+1)+x] (∞/∞)
=lim(x->∞) 1 / (x/√(x^2+1)+1 )
=...
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lim(x->∞)x[√(x^2+1)-x]
=lim(x->∞)x [√(x^2+1)-x][√(x^2+1)+x]/ [√(x^2+1)+x]
=lim(x->∞) x / [√(x^2+1)+x] (∞/∞)
=lim(x->∞) 1 / (x/√(x^2+1)+1 )
=lim(x->∞) √(x^2+1) / (x+√(x^2+1) )
=lim(x->∞) √(1+1/x^2) / (1+√(1+1/x^2) )
=1/2
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