Prove that e^x ﹥ 0 (ILIEKMATHPHYSICS)

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  • Опубликовано: 26 ноя 2024

Комментарии • 2

  • @goncalofreitas2094
    @goncalofreitas2094 2 часа назад +2

    Neat

  • @benjaminschmutter
    @benjaminschmutter 2 часа назад +1

    You can also pretty easily prove this by letting N > -x, and seeing that for all n > N, 1+x/n > 0, meaning that the limit of (1+x/n)^n > 0.

    • @VicTheMathMan
      @VicTheMathMan Час назад

      Remember that when we work with limits the inequalities situations has a different behavior. It's possible having a convergent sequence x_n with every term greater than a ( a is a real number) and its limit is equal or greater than a.
      Example : The sequence 1/n is strictly positive, although its limits is equal to 0

  • @saaah707
    @saaah707 2 часа назад +2

    e^x > 0, and then moment generating functions go brrrr
    consider covering some proofs from probability or statistical inference at some point