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hank612
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4Â¥2014-12-02 06:01:43
zaq123321
ר¼Ò¹ËÎÊ (ÖøÃûдÊÖ)
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lixuemei201(feixiaolin´ú·¢): ½ð±Ò+1 2014-12-02 12:37:38
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lixuemei201(feixiaolin´ú·¢): ½ð±Ò+1 2014-12-02 12:37:38
| It seems title is different with the question. Also, it seems the question is about to use Rolle's theorem and l'hopital's rule, but there is not condition that g(x) is of C^1[a,+\infty). |

2Â¥2014-12-02 03:38:23
Edstrayer
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3Â¥2014-12-02 04:04:35
zaq123321
ר¼Ò¹ËÎÊ (ÖøÃûдÊÖ)
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lixuemei201(feixiaolin´ú·¢): ½ð±Ò+1 2014-12-02 12:37:56
lixuemei201(feixiaolin´ú·¢): ½ð±Ò+1 2014-12-02 12:37:56
|
º¯Êýx£¨f£¨x£©-g£¨x£©)£¬µ±xÇ÷ÓÚÕýÎÞÇîʱ¼«ÏÞ´æÔÚ => f(x)-g(x)->0, x->+\infty. => f(x) ->g(x), x->+\infty. From f£¨x£©ÔÚ[a,£«00)Á¬Ðø¿Éµ¼£¬ÇÒÆäµ¼º¯ÊýÓнç, => f(x)->C, x->+\infty. ????, then g(x)->C, as x->+\infty. By ±ÕÇø¼äÉϵÄÁ¬Ðøº¯ÊýÒ»ÖÂÁ¬Ðø => g(x)Ò»ÖÂÁ¬Ðø |

5Â¥2014-12-02 06:15:11













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