16 дек. 2014 г. · Any ideas on how to calculate the limit of (ex−1)/x as x goes to zero without applying L'Hôpital's rule? calculus · limits · exponential- ... |
28 февр. 2017 г. · limx→0+xe1/x=0⋅∞ which is an indefinite form. We could apply L'Hopital Rule but it can only be done if we have a 0/ ... |
8 авг. 2018 г. · The 0+ is not a real number just means something tends to 0. So the answer is ∞ suggests e1x→∞ is faster than x→0+. |
16 июн. 2016 г. · Use the Taylor series of the exponential : ex=∞∑0xnn! limx→0ex−1x=limx→0(1+x+o(x)−1)x=limx→0x+o(x)x=1. You can also recognise the definition ... |
29 окт. 2019 г. · The definition of the number e that's used in my textbook is e=limx→∞(1+1x)x which relates to compound interest. But when trying to calculate ... |
31 мар. 2018 г. · You can compute the limit using ex−1=O(x). Contrary, the limit in the title is just a number (0). |
26 июл. 2014 г. · We can find the limit: limn→+∞n(e−(1+1n)n). by exploiting the convexity of the exponential function, for which: x<y⟹ex<ey−exy−x<ey. |
9 янв. 2017 г. · Note that we can write. (ex−1)1/x=e(1−e−x)1/x. The limit, limx→∞(1−e−x)1/x, is not of indeterminate form since 10=1. Therefore, we have. |
26 авг. 2018 г. · I recently came up with a problem which I think is quite interesting and here it is. Let I(n)=limx→0(1(ex−1)n−1(x1!+x22!+x33!+⋯+xnn!)n). |
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