I need a proof that lim(1+1/x)^x=e.

We defined that e=lim(1+1/n)^n and I need to prove that it's same for x (real number).

Thanks.

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- Apr 21st 2013, 05:06 AM #1

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- Apr 21st 2013, 05:39 AM #2
## Re: lim (1+1/x)^x

The question seems trivial, as:

$\displaystyle \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e$

Is the exact same statement as:

$\displaystyle \lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{x}=e$

Just let n = x.

$\displaystyle n \epsilon \mathbb{R}$

$\displaystyle x \epsilon \mathbb{R}$

Though, this also seems trivial.. Even if they were integers or natural numbers, it would still approach the value of 'e' as the limit went to infinity.

Have you done the proof for the 'n' in class?

- Apr 21st 2013, 06:14 AM #3
## Re: lim (1+1/x)^x

That is no proof. Real numbers are not all integers.

To do this we need to have two additional facts: assume that $\displaystyle x\ge 1$ then

$\displaystyle \\1)~\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n+1} \right )^{n}=e\\\\2)~\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n+1}=e$

Now use the floor function (*greatest*): $\displaystyle \left\lfloor x \right\rfloor \leqslant x < \left\lfloor x \right\rfloor + 1$.__integer__function

Thus $\displaystyle \left( {1 + \frac{1}{{\left\lfloor x \right\rfloor }}} \right) \geqslant \left( {1 + \frac{1}{x}} \right) \geqslant \left( {1 + \frac{1}{{\left\lfloor x \right\rfloor + 1}}} \right)$

So $\displaystyle {\left( {1 + \frac{1}{{\left\lfloor x \right\rfloor }}} \right)^{\left\lfloor x \right\rfloor + 1}} \geqslant {\left( {1 + \frac{1}{x}} \right)^x} \geqslant {\left( {1 + \frac{1}{{\left\lfloor x \right\rfloor + 1}}} \right)^{\left\lfloor x \right\rfloor }}$

- Apr 21st 2013, 07:07 AM #4
## Re: lim (1+1/x)^x

We know that:

$\displaystyle \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n} = \lim_{n\rightarrow \infty } e^{\ln{\left( 1+\frac{1}{n} \right)^{n}}}$

$\displaystyle \text{Let } N = \frac{1}{n}$

and as $\displaystyle n \rightarrow \infty $ so $\displaystyle N \rightarrow 0$

$\displaystyle \therefore \lim_{n\rightarrow \infty } e^{\ln{\left( 1+\frac{1}{n} \right)^{n}}} = \lim_{N\rightarrow 0 } e^{\ln{\left( 1+N \right)^{\frac{1}{N}}}}$

$\displaystyle \begin{align*}\lim_{N \rightarrow 0 } e^{\ln{(1+N)^{1/N}}} =& \lim_{N \rightarrow 0 } e^{\frac{\ln{(1+N)}}{N}}\\=& e^{\lim_{N \rightarrow 0} \frac{\ln{(1+N)}}{N}}.......[\text{by power rule of limit}]\end{align*}$

Using I'Hopital's rule:

$\displaystyle \lim_{N \rightarrow 0} \frac{\ln{(1+N)}}{N} = \lim_{N \rightarrow 0} \frac{\frac{1}{1+N}}{1} = 1$

$\displaystyle \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n} = e^{1} = e$

$\displaystyle \text{Q.E.D}$

- Apr 21st 2013, 07:20 AM #5
## Re: lim (1+1/x)^x

@x3bnm, Did you read the OP?

Had you, you would have noted that the fact that $\displaystyle \lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e$ was never in question.

In fact, the OP stipulates**that is the definition of**$\displaystyle e$.

The you are asked to show that $\displaystyle {\lim _{x \to \infty }}{\left( {1 + \frac{1}{x}} \right)^x} = e$.

- Apr 21st 2013, 07:23 AM #6