1. ## lim (1+1/x)^x

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.

2. ## Re: lim (1+1/x)^x

The question seems trivial, as:

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

Is the exact same statement as:

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

Just let n = x.
$n \epsilon \mathbb{R}$
$x \epsilon \mathbb{R}$

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?

3. ## Re: lim (1+1/x)^x

The question seems trivial, as:
$\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e$
Is the exact same statement as:
$\lim_{x\rightarrow \infty }\left ( 1+\frac{1}{x} \right )^{x}=e$
Just let n = x.
$n \epsilon \mathbb{R}$
$x \epsilon \mathbb{R}$
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.
That is no proof. Real numbers are not all integers.

To do this we need to have two additional facts: assume that $x\ge 1$ then
$\\1)~\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n+1} \right )^{n}=e\\\\2)~\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n+1}=e$

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

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

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

4. ## Re: lim (1+1/x)^x

We know that:

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

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

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

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

\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:

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

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

$\text{Q.E.D}$

5. ## Re: lim (1+1/x)^x

Originally Posted by x3bnm
$\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n} = e^{1} = e$

@x3bnm, Did you read the OP?
Had you, you would have noted that the fact that $\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e$ was never in question.
In fact, the OP stipulates that is the definition of $e$.

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

6. ## Re: lim (1+1/x)^x

Originally Posted by Plato
@x3bnm, Did you read the OP?
Had you, you would have noted that the fact that $\lim_{n\rightarrow \infty }\left ( 1+\frac{1}{n} \right )^{n}=e$ was never in question.
In fact, the OP stipulates that is the definition of $e$.

The you are asked to show that ${\lim _{x \to \infty }}{\left( {1 + \frac{1}{x}} \right)^x} = e$.
Sorry Plato. My bad. I thought(erroneously) he wanted to know why it's equal to $e$. Readers...please disregard my last post. Sorry.