f'(x)连续,f(0)=0,f'(x)不等于0,求lim（x趋于0）∫[from 0 to x^2]f(t)dt/{x
f'(x)连续,f(0)=0,f'(x)不等于0,求lim（x趋于0）∫[from 0 to x^2]f(t)dt/{x^2∫[from 0 to x]f(t)dt}
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lim（x趋于0）∫[from 0 to x^2]f(t)dt/{x^2∫[from 0 to x]f(t)dt}
=lim(x→0)2xf(x²)/{x²f(x)+2x∫[from 0to x ]f(t)dt}
=lim(x→0)2f(x²)/{xf(x)+2∫[from 0 to x]f(t) dt}
=lim(x→0)4xf'(x²)/{f(x)+xf '(x)+2f(x)}
=lim(x→0)4f'(x²)/[3f(x)/x+f'(x)]
因为lim(x→0)f(x)/x=lim(x→0)[f(x)-f(0)]/(x-0)=f'(0)
所以lim(x→0)4f'(x²)/[3f(x)/x+f'(x)]
=4f'(0)/[3f'(0)+f'(0)]
=1
答案：1
=lim(x→0)2xf(x²)/{x²f(x)+2x∫[from 0to x ]f(t)dt}
=lim(x→0)2f(x²)/{xf(x)+2∫[from 0 to x]f(t) dt}
=lim(x→0)4xf'(x²)/{f(x)+xf '(x)+2f(x)}
=lim(x→0)4f'(x²)/[3f(x)/x+f'(x)]
因为lim(x→0)f(x)/x=lim(x→0)[f(x)-f(0)]/(x-0)=f'(0)
所以lim(x→0)4f'(x²)/[3f(x)/x+f'(x)]
=4f'(0)/[3f'(0)+f'(0)]
=1
答案：1
1年前
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