this post was submitted on 11 Oct 2023
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submitted 1 year ago* (last edited 1 year ago) by hypertown@lemmy.world to c/memes@lemmy.ml
 
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[–] Ilovethebomb@lemmy.ml 3 points 1 year ago (5 children)
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[–] Ghyste@sh.itjust.works 3 points 1 year ago (8 children)
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[–] Haus@kbin.social 0 points 1 year ago* (last edited 1 year ago) (2 children)

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.

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[–] abcd@feddit.de -1 points 1 year ago (1 children)

Where are my programmer buddies? 🤘

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[–] EdanGrey@sh.itjust.works -3 points 1 year ago (2 children)
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