# Rigorous Prépa Math Pedagogy

The Classe Prépa Math for Grandes Écoles is uniquely French pedagogy – very rigorous based on solid abstract theories.

In this lecture the young French professor demonstrates how to teach students the rigorous Math à la Française:

$\displaystyle {\lim_{n\to\infty} \bigl( 1 + \frac{1}{n} \bigr)^{n} = e}$

1st Mistake:
$\displaystyle { \bigl( 1 + \frac{1}{n} \bigr) \xrightarrow [n\to\infty] {} 1 \implies \boxed {{\bigl( 1 + \frac{1}{n} \bigr)}^{n} \xrightarrow [n\to\infty] {} 1^{n} =1}}$(WRONG!!)

THEOREM 1 :
$\boxed { \text {If } {U}_{m} \xrightarrow [n\to\infty] {} \alpha \text{ and } f(x) \xrightarrow [x \to\alpha] {} \ell \text { then } f (U_{m}) \xrightarrow [n\to\infty] {} \ell}$

Note: The ‘x’ in f (x) is ${U}_{m}$ hence f is ‘fixed’ by a value $\alpha$

In the above mistake:
${U}_{m} = 1 + \frac{1}{n}$
$f _{n} (1 + \frac{1}{n} ) \text { where } f_{n} : x \mapsto x^{n}$
${f}_{n}$ is not fixed, but depends on ‘n’. It is wrong to apply Theorem 1.

2nd Mistake:
$\forall n \in \mathbb {N}^{*}, \bigl( 1 + \frac{1}{n} \bigr) > 1$

THEOREM 2:
$\boxed { \forall q > 1, q^{n} \xrightarrow [n\to +\infty] {} +\infty}$
Note: q is a fixed value.

$\text {Let } q_{n} = 1 + \frac{1}{n}$
Therefore,
$\boxed {{\bigl(1 + \frac{1}{n} \bigr)}^{n} \xrightarrow [n\to\infty] {} +\infty }$ (WRONG!!)

Reason: $q_{n} = \bigl( 1 + \frac{1}{n} \bigr)$ is not fixed value but depends on variable ‘n’. It is wrong to apply Theorem 2.

3rd Mistake: Binomial

$\boxed { \displaystyle \forall n \in \mathbb {N}^{*}, {\bigl( 1 + \frac{1}{n} \bigr)}^{n} = \sum_{k=0}^{n} \binom {n}{k} {(\frac {1}{n})}^{k} . (1)^{n-k} = \sum_{k=0}^{n} \binom {n}{k} {(\frac {1}{n})}^{k} }$

Note:
${\bigg[ \bigl( 1 + \frac{1}{n} \bigr)}^{n} = \binom {n}{0} {(\frac {1}{n})}^{0} + \binom {n}{1} {(\frac {1}{n})}^{1} + ... = 1 + 1 + ... > 2 \bigg]$
Expanding the binomial,
$\displaystyle \binom {n}{k} = \frac {n. (n-1). (n-2)... (n-k+1)}{k!} \sim_{n\to +\infty} \frac {n^k}{k!}$

$\boxed {\displaystyle\binom {n}{k}.{(\frac {1}{n})}^{k} \sim_{n\to +\infty}\frac {n^k}{k!}. {(\frac {1}{n})}^{k} =\frac {1}{k!}}$ Note: valid if k is fixed value.

$\boxed {\displaystyle {\bigl( 1 + \frac{1}{n} \bigr)}^{n} = \sum_{k=0}^{n} \binom {n}{k} {(\frac {1}{n})}^{k} \sim_{n\to +\infty} \sum_{k=0}^{n}\frac {1}{k!}}$ (WRONG !!)
Reason: k in the $\displaystyle \sum_{k=0}^{n}$ is not fixed, it varies from k = 0 to n

THEOREM 3:
$\displaystyle\boxed { {U}_{n} \sim_{n\to +\infty} {V}_{n}, \forall p \ge 1, ({U}_{n})^{p} \sim_{n\to +\infty} ({V}_{n})^{p} }$
Note: p is fixed value.
$\boxed {\displaystyle ({U}_{n})^{n} \sim_{n\to +\infty} ({V}_{n})^{n} }$ (WRONG!!)
Reason: n is variable to infinity.

Question:
$\text {If } {U}_{n}\xrightarrow [n\to\infty] {} 0, \text {then } {(1+{U}_{n})}^{\alpha} \xrightarrow [n\to\infty] {} \alpha.{U}_{n}$
Is below correct ?
${(1+\frac{1}{n})}^{n} \xrightarrow [n\to\infty] {??} n. \frac{1}{n} = 1$
[Hint] n is variable, $\alpha$ is fixed value.

Final Solution: Exponential

THEOREM 4:
$\boxed {\forall (a,b) \in \mathbb {R}_{+}^{*} \text { x }\mathbb {R}, a^{b} = e^{b.\ln {a}}}$

$\boxed { \displaystyle \forall n \in \mathbb {N}^{*}, {\bigl( 1 + \frac{1}{n} \bigr)}^{n} = e^{n. {\ln (1+ \frac {1}{n})}} }$ … [*]

THEOREM 5:
$\boxed {\text {If } {U}_{n}\xrightarrow [n\to\infty] {} 0, \text {then } \ln (1+{U}_{n}) \sim_{n\to\infty} {} {U}_{n} }$

Since
$\frac{1}{n} \xrightarrow [n\to\infty] {} 0, \text {then } \ln (1+ \frac{1}{n}) \sim_{n\to\infty} {} \frac{1}{n}$
Therefore,
$n.\ln (1+ \frac{1}{n}) \sim_{n\to\infty} {}1$

THEOREM 6: From equivalence ($\sim_{n\to\infty} {}$) to find Limit ($\xrightarrow [n\to\infty] {}$), and vice-versa
$\boxed { \text {If } {U}_{n} \sim_{n\to\infty} {} \ell \in \mathbb {R}, \text {then } {U}_{n} \xrightarrow [n\to\infty] {} \ell }$

Converse is true only if $\ell \neq 0$
eg. $\ell = (\frac {1}{n})_{n \in {N}^{*}} \neq 0$
because $(\frac {1}{n}) \xrightarrow [n\to\infty] {} 0$ but $\nsim_{n\to\infty} {} 0$

Therefore (from Theorem 6):
$\displaystyle \lim_{n\to\infty} n.\ln (1+ \frac{1}{n})= 1$

Since exponential function is continuous at 1 (why must state the condition of Continuity? )
hence, from [*], we have:
$\boxed { \displaystyle {\lim_{n\to\infty} \bigl( 1 + \frac{1}{n} \bigr)^{n} = e}}$