# Thread: Prove A Limit.

1. ## Prove A Limit.

Prove that:

(1 + x/n)^n = e^x as n approaches infinity

Without using l'Hopitals Rule?

The book proves (1 + 1/n)^n = e as n tends to infinity.
by considering f(x) = ln(x) and f'(x) = 1/x and f'(1) = 1
Then from first principles proves eventually that (1 + 1/n)^n = e as n tends to infinity, by letting n = 1/h etc.

2. Start by giving a name to your limit, say

$\displaystyle \varphi = \lim_{n \to \infty } \left( {1 + \frac{x} {n}} \right)^n.$

Since the logarithm is continuous on its domain, we can interchange the function and taking limits.

$\displaystyle \ln \varphi = \lim_{n \to \infty } n\ln \left( {1 + \frac{x} {n}} \right).$

Substitute $\displaystyle u=\dfrac1n,$

$\displaystyle \ln \varphi = \lim_{u \to 0} \frac{1} {u}\ln (1 + ux).$

Since $\displaystyle \ln (1 + ux) = \int_1^{1 + ux} {\frac{1} {v}\,dv} ,\,1 \le v \le 1 + ux,$

we have $\displaystyle \frac{1} {{1 + ux}} \le \frac{1} {v} \le 1\,\therefore \,\frac{x} {{1 + ux}} \le \frac{1} {u}\ln (1 + ux) \le x.$

Taking the limit as $\displaystyle u\to0,$ we conclude by Squeeze Theorem that $\displaystyle \lim_{u \to 0}\frac{1} {u}\ln (1 + ux) = x,$ and we happily get

$\displaystyle \ln \varphi = x\,\therefore \,\varphi = e^x .$

3. Originally Posted by Krizalid
Start by giving a name to your limit, say

$\displaystyle \varphi = \lim_{n \to \infty } \left( {1 + \frac{x} {n}} \right)^n.$
No! That is not good. You are assuming the limit exists. How do you know the limit exists?

What you should do is write,
$\displaystyle \left( 1 + \frac{x}{n} \right)^n = \exp \left[ n\ln \left( 1 + \frac{x}{n} \right) \right]$

Since $\displaystyle \ln (1 + ux) = \int_1^{1 + ux} {\frac{1} {v}\,dv} ,\,1 \le v \le 1 + ux,$

we have $\displaystyle \frac{1} {{1 + ux}} \le \frac{1} {v} \le 1\,\therefore \,\frac{x} {{1 + ux}} \le \frac{1} {u}\ln (1 + ux) \le x.$
Here you are assuming that $\displaystyle x>0$. It does not cover the case when $\displaystyle x<0$.
(The way I have it above also does $\displaystyle x>0$).

Anyway, good job.