lim(xto0)2[(e^x-(1+x))/x^2]=

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lim(xto0)2[(e^x-(1+x))/x^2]=

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\lim_{x\to 0} e^x-\frac{(1+x)}{x^2}\] ?
satellite u lost 2
ooh i know \[lim_{x\to 0} \frac{2e^x-(1+x)}{x^2}\]

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i bet that is it. l'hopital's rule problem. of the form \[\frac{0}{0}\]
take derivative top and bottom separate. you will need to do it twice because after the first time you still get \[\frac{0}{0}\]
the top is 2e^0-1-0=2(1)-1-0=2-1=1
first time \[\frac{2e^x-1}{2x}\]
still \[\frac{0}{0}\]
second time \[\frac{2e^x}{2}\] let x ->0 get 1
oops i am totally wrong. sorry. first time it is \[\frac{1}{0}\] no limit. so sorry.
must be my bed time.
yes you have lost the derivative of constant 1 first time
by first time i mean before you start. it is not \[\frac{0}{0}\] to begin with so no l'hopital. sorry
WHERE IS THE DERIVATIVE OF CONSTANT?
Mmmm....looking carefully at the question as posted, we are looking for the limit of 2(e^x - 1 - x)/x^2. As x goes to zero we get 2(e^0 - 1 - 0)/0^2 = 2(1 - 1)/0 = 0/0, so use l'Hopital twice to get Lim as x goes to 0 of e^x, which is 1.
did you try graphing it? both sides of 0 f(x) goes to positive infinity
WHY WE TAKE DERIVATIVE ANY REASON????

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