• Haus@kbin.social
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    1 year ago

    It’s been a while, but I think I remember this one. Lim 1/n =0 as n approaches infinity. Let x^0 be undefined. For any e>0 there exists an n such that |x^(1/n) -1| < e. If you desire x^(1/n) to be continuous at 0, you define x^0 as 1.

    E2a: since x^(1/n)>1, you can drop the abs bars. I think you can get an inequality to pick n using logs.

    • uberrice@feddit.de
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      1 year ago

      Simpler: x^1 = x, x^-1 = 1/x

      x^1 * x^-1 = x^0 = x/x = 1.

      Of course, your explanation is the “correct” one - why it’s possible that x^0=1. Mine is the simple version that shows how logic checks out using algebraic rules.

    • milicent_bystandr@lemm.ee
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      1 year ago

      What?! Impossible to start a family at 18 and also enjoy mathematics?

      Not everyone who has unprotected sex at 18 (or with an 18 yr old) is some numbskull just going at it for unscrupulous pleasure.

      (As another reply also pointed out: the pun was crafted by the OP’s dad, not the 1yr-old’s dad; and OP could be the child’s mum or dad)