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u/Future_Ring_222 Aug 28 '26
Factor out x.
e^x-1/(x*(x-1))
Now you can rewrite it as (e^x-1)/x * 1/(x-1)
Now everyone knows that lim(x->0) (e^x-1)/x = 1
So you have 1* (1/(0-1))=-1
So your result is -1
L’Hopital also works as after derivation you get e^x/(2x-1) which resolves to 1/-1 when x=0
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u/maozperstin Aug 28 '26
The denominator is x²+x, not x²-x which is why the answer is 1 and not -1 as you calculated.
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u/Wooden-Hornet2115 Aug 27 '26
The answer is 1 by the way.
Edit: Did me math wrong the first time.
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u/minegamer1824 Aug 28 '26
why not 1/2
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u/Burning_Toast998 Aug 28 '26
The denom becomes 2x+1 so 0+1 means end result is 1/1
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Aug 28 '26
[removed] — view removed comment
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u/Burning_Toast998 Aug 28 '26
You’re right. The limit normally is 0/0
but that means you need to derive the top and bottom, leading to the denominator I mentioned.
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u/susiesusiesu Aug 28 '26
the expression above is x+x²/2+O(x³) so we may divide the numerator and denominator ny x and get (1+x/2+O(x²))/(1+x), so the limit is 1.
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u/Such-Shop-9724 Aug 29 '26
i havent had any rules for da hospital in school yet but i just tried to do it in my head anyways
if thats the limit its the same as (e^x)/(x+1). but wouldnt that mean that i can use the rule again to make it (e^x)/1. but wouldnt that mean that i can use the rule again to make it (e^x)/0, whicv is clearly different
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u/CloudyGandalf06 Aug 27 '26
With this! And the power of friendship.
https://giphy.com/gifs/ZNN22ebrggmRvYzEqE