求教几题高数1.f(x)在x=5上连续且,在x无穷接近5的情况下,lim [f(x)-5]/(x-5)=8,求f(5)2.lim x→0 ln(1+f(x))/x^2=2 求lim x→0 f(x)/x^2=3lim n趋于无穷 sin(sqr(n^2+1)π)=
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![求教几题高数1.f(x)在x=5上连续且,在x无穷接近5的情况下,lim [f(x)-5]/(x-5)=8,求f(5)2.lim x→0 ln(1+f(x))/x^2=2 求lim x→0 f(x)/x^2=3lim n趋于无穷 sin(sqr(n^2+1)π)=](/uploads/image/z/13294078-70-8.jpg?t=%E6%B1%82%E6%95%99%E5%87%A0%E9%A2%98%E9%AB%98%E6%95%B01.f%28x%29%E5%9C%A8x%3D5%E4%B8%8A%E8%BF%9E%E7%BB%AD%E4%B8%94%2C%E5%9C%A8x%E6%97%A0%E7%A9%B7%E6%8E%A5%E8%BF%915%E7%9A%84%E6%83%85%E5%86%B5%E4%B8%8B%2Clim+%5Bf%EF%BC%88x%EF%BC%89-5%5D%2F%EF%BC%88x-5%EF%BC%89%3D8%2C%E6%B1%82f%EF%BC%885%292.lim+x%E2%86%920+ln%281%2Bf%28x%29%29%2Fx%5E2%3D2+%E6%B1%82lim+x%E2%86%920+f%EF%BC%88x%EF%BC%89%2Fx%5E2%3D3lim+n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7+sin%28sqr%28n%5E2%2B1%29%CF%80%29%3D)
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求教几题高数1.f(x)在x=5上连续且,在x无穷接近5的情况下,lim [f(x)-5]/(x-5)=8,求f(5)2.lim x→0 ln(1+f(x))/x^2=2 求lim x→0 f(x)/x^2=3lim n趋于无穷 sin(sqr(n^2+1)π)=
求教几题高数
1.f(x)在x=5上连续且,在x无穷接近5的情况下,lim [f(x)-5]/(x-5)=8,求f(5)
2.lim x→0 ln(1+f(x))/x^2=2 求lim x→0 f(x)/x^2=
3lim n趋于无穷 sin(sqr(n^2+1)π)=
求教几题高数1.f(x)在x=5上连续且,在x无穷接近5的情况下,lim [f(x)-5]/(x-5)=8,求f(5)2.lim x→0 ln(1+f(x))/x^2=2 求lim x→0 f(x)/x^2=3lim n趋于无穷 sin(sqr(n^2+1)π)=
1.由于连续 若f(x)≠5 则lim(f(x)-5)/(x-5)=∞显然不行
所以x=5 至于8 运用洛必达法则只能看出这是f(x)在x=5时斜率
2.lim(f(x)/x^2)=lim(e^(ln(1+f(x)))-1)/(x^2)
洛必达法则--〉lim(e^(ln(1+f(x)))*(1/(1+f(x))*f'(x))/(2x)
又前者洛必达法则可得lim(1/(1+fx)*f'(x))/x^2=2
-->带入得lim(e^(ln(1+f(x)))*2=2
3.好像没有极限吧……
求教几题高数1.f(x)在x=5上连续且,在x无穷接近5的情况下,lim [f(x)-5]/(x-5)=8,求f(5)2.lim x→0 ln(1+f(x))/x^2=2 求lim x→0 f(x)/x^2=3lim n趋于无穷 sin(sqr(n^2+1)π)=
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