求lim x→1 2x-3/x^2-5x+4 因为lim x→1 x^2-5x+4/2x-3 = 1^2-5×1+4/2×1-3 = 0,根据无穷大与求lim x→1 2x-3/x²-5x+4 因为lim x→1 x²-5x+4/2x-3 = 1²-5×1+4/2×1-3 = 0,根据无穷大与无穷小的关系得lim x→1 2x-3/x
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![求lim x→1 2x-3/x^2-5x+4 因为lim x→1 x^2-5x+4/2x-3 = 1^2-5×1+4/2×1-3 = 0,根据无穷大与求lim x→1 2x-3/x²-5x+4 因为lim x→1 x²-5x+4/2x-3 = 1²-5×1+4/2×1-3 = 0,根据无穷大与无穷小的关系得lim x→1 2x-3/x](/uploads/image/z/3758712-24-2.jpg?t=%E6%B1%82lim+x%E2%86%921+2x-3%2Fx%5E2-5x%2B4+%E5%9B%A0%E4%B8%BAlim+x%E2%86%921+x%5E2-5x%2B4%2F2x-3+%3D+1%5E2-5%C3%971%2B4%2F2%C3%971-3+%3D+0%2C%E6%A0%B9%E6%8D%AE%E6%97%A0%E7%A9%B7%E5%A4%A7%E4%B8%8E%E6%B1%82lim+x%E2%86%921+2x-3%2Fx%26%23178%3B-5x%2B4+%E5%9B%A0%E4%B8%BAlim+x%E2%86%921+x%26%23178%3B-5x%2B4%2F2x-3+%3D+1%26%23178%3B-5%C3%971%2B4%2F2%C3%971-3+%3D+0%2C%E6%A0%B9%E6%8D%AE%E6%97%A0%E7%A9%B7%E5%A4%A7%E4%B8%8E%E6%97%A0%E7%A9%B7%E5%B0%8F%E7%9A%84%E5%85%B3%E7%B3%BB%E5%BE%97lim+x%E2%86%921+2x-3%2Fx%26%231)
求lim x→1 2x-3/x^2-5x+4 因为lim x→1 x^2-5x+4/2x-3 = 1^2-5×1+4/2×1-3 = 0,根据无穷大与求lim x→1 2x-3/x²-5x+4 因为lim x→1 x²-5x+4/2x-3 = 1²-5×1+4/2×1-3 = 0,根据无穷大与无穷小的关系得lim x→1 2x-3/x
求lim x→1 2x-3/x^2-5x+4 因为lim x→1 x^2-5x+4/2x-3 = 1^2-5×1+4/2×1-3 = 0,根据无穷大与
求lim x→1 2x-3/x²-5x+4 因为lim x→1 x²-5x+4/2x-3 = 1²-5×1+4/2×1-3 = 0,根据无穷大与无穷小的关系得lim x→1 2x-3/x²-5x+4 = ∞ .我想问一下为什么第一步就把分子分母倒过来,有什么依据吗?还有,为什么最后的结果是无穷大?算出来的不是0吗?如果有牵扯到得公式定理请写一下.
求lim x→1 2x-3/x^2-5x+4 因为lim x→1 x^2-5x+4/2x-3 = 1^2-5×1+4/2×1-3 = 0,根据无穷大与求lim x→1 2x-3/x²-5x+4 因为lim x→1 x²-5x+4/2x-3 = 1²-5×1+4/2×1-3 = 0,根据无穷大与无穷小的关系得lim x→1 2x-3/x
因为原式的分母在x→1时为0,分子为常数;而倒过来的话,分母就不为零了.算出来的是原极限的倒数,根据无穷小的倒数为无穷大,结果当然为无穷大了.
lim x→0 1/x=∞
方法是对的,你不是倒了后结果是0,所以原式的结果就是无穷大,就比如1/x趋近于0,所以x趋近于无穷