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}}

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