作业帮limx[(x^2+100)^1/2+x] (x->∞)题目中(x^2+100)^1/2和x是分开的两项相加x接近负无穷
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![limx[(x^2+100)^1/2+x] (x->∞)题目中(x^2+100)^1/2和x是分开的两项相加x接近负无穷](/uploads/image/z/12495287-47-7.jpg?t=limx%5B%28x%5E2%2B100%29%5E1%2F2%2Bx%5D+%28x-%3E%E2%88%9E%29%E9%A2%98%E7%9B%AE%E4%B8%AD%28x%5E2%2B100%29%5E1%2F2%E5%92%8Cx%E6%98%AF%E5%88%86%E5%BC%80%E7%9A%84%E4%B8%A4%E9%A1%B9%E7%9B%B8%E5%8A%A0x%E6%8E%A5%E8%BF%91%E8%B4%9F%E6%97%A0%E7%A9%B7)
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limx[(x^2+100)^1/2+x] (x->∞)题目中(x^2+100)^1/2和x是分开的两项相加x接近负无穷
limx[(x^2+100)^1/2+x] (x->∞)
题目中(x^2+100)^1/2和x是分开的两项相加
x接近负无穷
limx[(x^2+100)^1/2+x] (x->∞)题目中(x^2+100)^1/2和x是分开的两项相加x接近负无穷
只要把分子乘以[(x^2+100)^1/2-x]就可以,这样分子就会平方差相减,分子为100X,分母为[(x^2+100)^1/2-x],由于X是负无穷,所以(x^2+100)^1/2就近似为绝对值X ,即负X,所以分母为-2X,那么100X/-2X=-50,