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I am confused as to why is it that only e is such that d/dx(e^x)=e^x

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I am confused as to why is it that only e is such that d/dx(e^x)=e^x

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pwsnafu

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We can give you mathematical proofs, but it seems like in your case, you shouldWhy is it that the constant e is defined as being the unique case where the limit as h goes to 0 of (e^h - 1)/h = 1?

a=0I mean every exponential function like a^x equals 1 when x equals 0, right?

But do they approachSo would it be fair to say that (a^h) approaches 1 as h approaches zero? And that (a^h - 1) approaches zero as h approaches zero? And finally that the limit as h goes to 0 of (a^h - 1)/h also equals one?

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I mean every exponential function like a^x equals 1 when x equals 0, right?

aa=0

To the OP, try taking y = a

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Also, is rate the only thing that matters? So if I had two things approaching zero at the same rate, but one of them would reach zero earlier because it starts lower, would the ratio of those two things still be equal to one?

And if that is not the case, does that mean that (e^h - 1) = h as h approaches zero?

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HallsofIvy

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For example, if we take h= .1, .01, .001, successively, [itex]2^h- 1[/itex] becomes

0.07177, 0.00695, and .00069. Each of those, divided by h, is 0.7177, 0.695, and 0.690, respectively. Each of those is less than 1 and it can be shown that they converge to a number less than 1.

But [itex]3^h- 1[/itex] becomes 0.11612, 0.01104, and 0.00109. Again dividing by h they give 1.1612, 1.104, and 1.09. Each of those is larger than 1 and converges to a number larger than 1.

We could do the same with, say, [itex](2.7^h- 1)/h[/itex] and [itex](2.75^h- 1)/h[/itex] showing that the first converges to a number slightly less than 1, the second to a number just larger than 1. There exist a between 2 and 3, between 2.7 and 2.75, such that [itex](a^h- 1)/h[/itex] converges to 1. We call that value, "e".

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