f(x)在[a,b]上连续,(a,b)上可导,且f′(x)>0,若x趋向于a+,limf(2x-a)/(x-a)存在,证明:在(a,b)内,f(x)>0
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![f(x)在[a,b]上连续,(a,b)上可导,且f′(x)>0,若x趋向于a+,limf(2x-a)/(x-a)存在,证明:在(a,b)内,f(x)>0](/uploads/image/z/8709833-65-3.jpg?t=f%28x%29%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%28a%2Cb%29%E4%B8%8A%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%E2%80%B2%28x%29%3E0%2C%E8%8B%A5x%E8%B6%8B%E5%90%91%E4%BA%8Ea%2B%2Climf%282x-a%29%2F%28x-a%29%E5%AD%98%E5%9C%A8%2C%E8%AF%81%E6%98%8E%3A%E5%9C%A8%28a%2Cb%29%E5%86%85%2Cf%28x%29%3E0)
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f(x)在[a,b]上连续,(a,b)上可导,且f′(x)>0,若x趋向于a+,limf(2x-a)/(x-a)存在,证明:在(a,b)内,f(x)>0
f(x)在[a,b]上连续,(a,b)上可导,且f′(x)>0,若x趋向于a+,limf(2x-a)/(x-a)存在,证明:在(a,b)内,f(x)>0
f(x)在[a,b]上连续,(a,b)上可导,且f′(x)>0,若x趋向于a+,limf(2x-a)/(x-a)存在,证明:在(a,b)内,f(x)>0
由于x趋于a+时,分母x-a是趋于0的,所以如果极限limf(2x-a)/(x-a)存在,分子f(2x-a)也必须趋于0,这样的0/0型未定式极限才可能存在.故x趋于a+时有limf(2x-a)=0,由于f(x)在x=a+处连续,故limf(2x-a)=f(2a-a)=f(a),由此可知f(a)=0,而在(a.b)内f'(x)>0,即f(x)严格递增,所以有f(x)>f(a)=0,这样就证明了在(a,b)内f(x)>0.
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