高数求极限问题,下图里这三道题解答过程都看不懂,麻烦高手讲解一下,最好详细点,谢谢啦! 5
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(6)
lim(x->α) (sinx- sinα) /(x-α) (0/0)
=lim(x->α) cosx
=cosα
(7)
√(x^2+x) - √(x^2-x)
=[√(x^2+x) - √(x^2-x)] . [√(x^2+x) + √(x^2-x)]/[√(x^2+x) + √(x^2-x)]
= 2x/[√(x^2+x) + √(x^2-x)]
lim(x->∞) [√(x^2+x) - √(x^2-x) ]
=lim(x->∞) 2x/[√(x^2+x) + √(x^2-x)]
=lim(x->∞) 2/[√(1+1/x) + √(1-1/x)]
=2/(1+1)
=1
(8)
lim(x->0) [ ( 1- (1/2)x^2)^(2/3) -1 ]/[xln(1+x) ]
=lim(x->0) - (1/3)x^2 /x^2
=-1/3
x->0
( 1- (1/2)x^2)^(2/3) ~ 1 - (2/3)(1/2)x^2 = 1- (1/3)x^2
( 1- (1/2)x^2)^(2/3) -1 ~ - (1/3)x^2
ln(1+x) ~ x
xln(1+x) ~ x^2
lim(x->α) (sinx- sinα) /(x-α) (0/0)
=lim(x->α) cosx
=cosα
(7)
√(x^2+x) - √(x^2-x)
=[√(x^2+x) - √(x^2-x)] . [√(x^2+x) + √(x^2-x)]/[√(x^2+x) + √(x^2-x)]
= 2x/[√(x^2+x) + √(x^2-x)]
lim(x->∞) [√(x^2+x) - √(x^2-x) ]
=lim(x->∞) 2x/[√(x^2+x) + √(x^2-x)]
=lim(x->∞) 2/[√(1+1/x) + √(1-1/x)]
=2/(1+1)
=1
(8)
lim(x->0) [ ( 1- (1/2)x^2)^(2/3) -1 ]/[xln(1+x) ]
=lim(x->0) - (1/3)x^2 /x^2
=-1/3
x->0
( 1- (1/2)x^2)^(2/3) ~ 1 - (2/3)(1/2)x^2 = 1- (1/3)x^2
( 1- (1/2)x^2)^(2/3) -1 ~ - (1/3)x^2
ln(1+x) ~ x
xln(1+x) ~ x^2
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