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You don’t need l’hôpital’s rule for this. Using it to prove the derivative of sin x is cos x is almost backwards or circular logic. By definition of derivative you have already needed to know the limit as x approaches 0 before you could even differentiate sin x.
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Am I being stupid here - how is 1^∞ indeterminate? Maybe displaying my inner physicist but isn't it.... 1?
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Depends, when I had to study for my uni exams a case we characterized as 1 to infinity is the lim (1+1/n)^n which is the definition of e. You cannot simply say oh 1+1/n -> 1 as n approaches infinity so the whole limit is converging to 1. I do not know if this helps.
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Not sure about 1^inf ? Surely that's 1 ? I'd be interested to know why not
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Sorry still don't get it 1^1 = 1 1^2 = 1 1^3 = 1 1^4 = 1 .. 1^inf = 1 surely?
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L’HOPITAL
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Niles Crepney of East Doorflap (1896-1962) approached his limit when his dim wife Claire brought home 4 stone of smelt from Stebbins’ Market. “Don’t much like ‘em, all the bones an’ all,” he complained. Settled for a nice bubble and squeak at Dolly’s Pub.
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the multiplication operation has no defined output for any situation where an operand is undefined. it’s tempting to use the properties of zero to ignore this, but 0 x undefined is undefined.
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Could someone enlighten me please? Why is zero times infinity indeterminate?
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Because you cannot determinate the value of such expressions just by looking at the limits. For example, both exp(-x)*x and 1/x * exp(x) are of the form 0*∞, but the first one goes to 0 and the second one goes to ∞ (as x→∞).
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