Question

Prove that: \[
\lim_{x\to 0} \frac{a^x-1}{x} = 1 \iff \lim_{x\to \infty} \left(1+\frac 1x\right)^x=a
\]

Answer 1

Math Battle!

First person to prove correctly deserves 20 medals.

Answer 2

We have to prove that a=1? or what?

Answer 3

No, you have to prove that \(a\) satisfies the first limit if an only if it is the result of the second limit.

Answer 4

Have to prove both directions, that if the LHS holds then the RHS follows, and vice versa.

Answer 5

\[ \lim_{x\to \infty} \left(1+\frac 1x\right)^x=a\]
\[\LARGE \lim_{x\to \infty} \left(1+\frac 1x\right)^{x \times \frac{x}{x}}=e\]
isn't it "e"? how did a come there o.O?

Answer 6

Knowing that \(a=\exp(1)\) will not help you here.

Answer 7

\(\lim_{x\to 0} \frac{a^x-1}{x} = 1 = \ln e \)

Answer 8

=> a= e

Answer 9

STOP with "e", people!! :) The point is NOT to relate the limit to e. The point is to prove that:

If LHS holds, then the RHS holds....
and
if RHS holds, then the LHS holds.

Does not matter if we all know that you can substitute e for a, that doesn't prove what is asked. You are just using something you already know, not proving what is here. :)

Answer 10

Again, knowing that will not help you hear. You can assume either definition of \(e\), but you can't assume both. Otherwise you are assuming the the proposition is true. That can't be done in a proof.

Answer 11

why not, prove LHS = 1 only when a=e
then prove that RHS = a, only when LHS holds

Answer 12

that should okay ofr a proof in one direction,
we can do the same in other direction

Answer 13

Because you are relying on two definitions for the same concept.

Answer 14

You can't assume the proposition it already true as a proof.

Answer 15

nope, we can prove LHS = ln a right ?

Answer 16

\[
|x|<\delta \implies \left|\frac{a^x-1}{x}-1\right|<\epsilon
\]

Answer 17

\[
x>\delta '\implies \left|\left(1+\frac{1}{x}\right)^x-a\right|<\epsilon'
\]

Answer 18

@DebbieG I should have just wrote these definitions so that they wouldn't start pulling the lazy tactics!

Answer 19

\[\lim_{x\to 0} \frac{a^x-1}{x} = 1 \iff \lim_{x\to \infty} \left(1+\frac 1x\right)^x=a\]

Assume that:
\[\lim_{x\to 0} \frac{a^x-1}{x} = 1\]
L'hospital's rule applies, so take derivatives:

\[\lim_{x\to 0} \frac{a^x\cdot ln(a)}{1} = 1\]

\[ ln(a)\cdot\lim_{x\to 0} \frac{a^x}{1} = 1\]
\[ ln(a) = 1\]
so a=e.

NOW, can we use the known result that the RHS limit is =e?

Answer 20

No ⒶArchie☁✪, we have to prove that IF LHS holds, then RHS follows; and also, IF RHS hold, then LHS follows.

It isn't a proof of either side's statement; is is the proof that each statement implies the other is true.

Answer 21

But I'm not sure if we can use, in the proof, the fact that the RHS limit is =e.... lol.

Answer 22

@DebbieG Your proof assumes that \(\frac{d}{dx}e^x=e^x\)

Answer 23

Hmmm. yes, I guess it does.

Answer 24

So, this is a biconditional. Meaning we have to prove both directions...
so, for our first part, we will assume it is true and then try to prove the second part.

Answer 25

Which assume the right hand definition.

Which is perfectly fine, but you can't then claim the left hand definition.

Answer 26

Is it just a coincidence that \(a\) fits both limits? It's very interesting, but not intuitive.