limit of infinity(3題)高手進

http://i32.photobucket.com/albums/d50/pzsing/q1.jpg

http://i32.photobucket.com/albums/d50/pzsing/q2.jpg

http://i32.photobucket.com/albums/d50/pzsing/q3.jpg

1

lim( n→+∞.) (1+1/2+...+1/2^n)
=lim( n→+∞.) [1-(1/2)^n]/(1-1/2)
=lim( n→+∞.) 2[1-(1/2)^n]
=2*1
=2

2

(x^3+x)^(1/3)-(x^3+3)^(1/3)
=(x-3)/[(x^3+x)^(2/3)+(x^3+x)^(1/3)(x^3+3)^(1/3)+(x^3+3)^(2/3)]
=(1-3/x)/[(x^3+x)^(2/3)+(x^3+x)^(1/3)(x^3+3)^(1/3)+(x^3+3)^(2/3)]/x

Consider
(x^3+x)^(2/3)/x
=(x^2+1)/x^(1/3)
→+∞ as x→+∞

So
lim (x→+∞)(x^3+x)^(1/3)-(x^3+3)^(1/3)
=1/+∞
=0

3
lim (x→0)3x/(sinx-x)
=lim (x→0)3/(sinx/x-1)

since lim (x→0)(sinx/x-1)=1-1=0

The originial limit does not exist

2008-04-15 13:14:54 補充：
a^3-b^3=(a-b)(a^2+ab+b^2)

Of course you should know it already !!