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Gerry Myerson
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Re-opened.

Please reopen:

A Basic Limit From Exponentials

I believe this is not a duplicate of How does one prove that $e$ exists? which was why it was closed.

The question from the first link asks why the limit $\ \displaystyle\lim_{x \to 0} \frac{2^x-1}{x}\ $ exists. Whereas the question from the second link asks to prove that there exists a number $a$ such that $\ \displaystyle\lim_{h \to 0} \frac{a^h - 1}{h} = 1.$

Please reopen:

A Basic Limit From Exponentials

I believe this is not a duplicate of How does one prove that $e$ exists? which was why it was closed.

The question from the first link asks why the limit $\ \displaystyle\lim_{x \to 0} \frac{2^x-1}{x}\ $ exists. Whereas the question from the second link asks to prove that there exists a number $a$ such that $\ \displaystyle\lim_{h \to 0} \frac{a^h - 1}{h} = 1.$

Re-opened.

Please reopen:

A Basic Limit From Exponentials

I believe this is not a duplicate of How does one prove that $e$ exists? which was why it was closed.

The question from the first link asks why the limit $\ \displaystyle\lim_{x \to 0} \frac{2^x-1}{x}\ $ exists. Whereas the question from the second link asks to prove that there exists a number $a$ such that $\ \displaystyle\lim_{h \to 0} \frac{a^h - 1}{h} = 1.$

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Adam Rubinson
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Please reopen:

A Basic Limit From Exponentials

I believe this is not a duplicate of How does one prove that $e$ exists? which was why it was closed.

The question from the first link asks why the limit $\ \displaystyle\lim_{x \to 0} \frac{2^x-1}{x}\ $ exists. Whereas the question from the second link asks to prove that there exists a number $a$ such that $\ \displaystyle\lim_{h \to 0} \frac{a^h - 1}{h} = 1.$