lim(x->0)(1-e^1/x)/(1+e^1/x)不存在,此题如何解释左右极限不等?证明lim(x->0)(1-e^1/x)/(1+e^1/x)不存在证:原式=lim(x->0){[2-1-e^(1/x)]/[1+e^(1/x)]}=lim(x->0){2/[1+e^(1/x)]-1}∵右极限=lim(x->0+){2/[1+e^(1/x)]-1}=-1左极限=lim
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/28 07:09:17
![lim(x->0)(1-e^1/x)/(1+e^1/x)不存在,此题如何解释左右极限不等?证明lim(x->0)(1-e^1/x)/(1+e^1/x)不存在证:原式=lim(x->0){[2-1-e^(1/x)]/[1+e^(1/x)]}=lim(x->0){2/[1+e^(1/x)]-1}∵右极限=lim(x->0+){2/[1+e^(1/x)]-1}=-1左极限=lim](/uploads/image/z/7104944-56-4.jpg?t=lim%28x-%3E0%29%281-e%5E1%2Fx%29%2F%281%2Be%5E1%2Fx%29%E4%B8%8D%E5%AD%98%E5%9C%A8%2C%E6%AD%A4%E9%A2%98%E5%A6%82%E4%BD%95%E8%A7%A3%E9%87%8A%E5%B7%A6%E5%8F%B3%E6%9E%81%E9%99%90%E4%B8%8D%E7%AD%89%3F%E8%AF%81%E6%98%8Elim%28x-%3E0%29%281-e%5E1%2Fx%29%2F%281%2Be%5E1%2Fx%29%E4%B8%8D%E5%AD%98%E5%9C%A8%E8%AF%81%EF%BC%9A%E5%8E%9F%E5%BC%8F%3Dlim%28x-%3E0%29%7B%5B2-1-e%5E%281%2Fx%29%5D%2F%5B1%2Be%5E%281%2Fx%29%5D%7D%3Dlim%28x-%3E0%29%7B2%2F%5B1%2Be%5E%281%2Fx%29%5D-1%7D%E2%88%B5%E5%8F%B3%E6%9E%81%E9%99%90%3Dlim%28x-%3E0%2B%29%7B2%2F%5B1%2Be%5E%281%2Fx%29%5D-1%7D%3D-1%E5%B7%A6%E6%9E%81%E9%99%90%3Dlim)
xTN@A`O։XTjQT6H)M! *TVT(K43vVBg5UYzs=&:2F+KX[RX>QFsǣ5zgG%oˀ|d np@
:(e%BT]B7bt፡u%+0!.0}˼zYϸ藺p0z#%_sgesEK484q8
M7'10{oj-~_XQ]m1'/ QD&H9vZ)Rxf力 ڢq: iqZq7Y$Zc1