Exercice 03-06
Soient \(a,b\in \mathbb{R}\), \(c,d\gt 0\), et soit \(x_n:= \frac{an+b}{cn+d}\). À l'aide de la définition de limite, montrer que
  1. Si \(a=2, b=-3, c=3, d=7\), alors \[ \lim_{n\to \infty} x_n=\frac{2}{3}\,. \]
  2. En général, quels que soient \(a,b,c,d\), \[ \lim_{n\to \infty} x_n=\frac{a}{c}\,. \]

il faut commencer par écrire explicitement ce que vaut la distance \[ |x_n-\tfrac23| =\Bigl| \tfrac{2n-3}{3n+7}-\tfrac{2}{3} \Bigr| \]

il n'y a à peu près aucune différence: on refait le même calcul avec des nombres \(a,b,c,d\) quelconques.

  1. Dans le cas particulier, commençons par étudier la différence \[ |x_n-\tfrac23| =\Bigl| \tfrac{2n-3}{3n+7}-\tfrac{2}{3} \Bigr|= \tfrac{23}{3(3n+7)}\,. \] Fixons un \(\varepsilon\gt 0\). On a \[\begin{aligned} |x_n-\tfrac23|\leqslant \varepsilon &\quad\Leftrightarrow\quad \tfrac{23}{3(3n+7)}\leqslant \varepsilon\\ &\quad\Leftrightarrow\quad n\geqslant \tfrac{1}{3}(\tfrac{23}{3\varepsilon}-7)\,. \end{aligned}\] Si on prend n'importe quel entier \(N\) plus grand que \(\frac{1}{3}(\tfrac{23}{3\varepsilon}-7)\), par exemple \[N:=\lfloor\tfrac{1}{3}(\tfrac{23}{3\varepsilon}-7)\rfloor+1\,,\] on a bien que \(|x_n-\frac23|\leqslant \varepsilon\) pour tout \(n\geqslant N\). Ceci montre donc bien que \(x_n\to \frac23\).

    (Sans aucune modification de ma part):

    To rigorously understand why \[ \lim_{n \to \infty} x_n = \frac{2}{3}, \] where \(x_n = \frac{2n - 3}{3n + 7}\), we'll use the formal \(\varepsilon\)-\(\delta\) definition of the limit of a sequence.

    Intuitive Understanding:
    First, let’s start with some intuition:
    • The sequence \(x_n = \frac{2n - 3}{3n + 7}\) consists of rational expressions where both the numerator and denominator are linear in \(n\).
    • As \(n\) becomes very large, the lower-order terms (constants like \(-3\) and \(7\)) in both the numerator and denominator become insignificant compared to the terms involving \(n\). This suggests that the behavior of \(x_n\) for large \(n\) is dominated by the ratio of the leading coefficients, which are \(2\) in the numerator and \(3\) in the denominator.
    • Therefore, for large \(n\), we expect \(x_n\) to behave approximately like \(\frac{2n}{3n} = \frac{2}{3}\).
    But now, let’s rigorously show this using the formal definition of a limit.
    Formal Definition of Limit of a Sequence:
    The formal definition of a limit of a sequence says that the sequence \((x_n)\) converges to \(L\) (in this case \(L = \frac{2}{3}\)) if, for every \(\varepsilon \gt 0\), there exists a positive integer \(N\) such that for all \(n \geqslant N\), \[ |x_n - L| \lt \varepsilon. \] In our case, we need to show that for any \(\varepsilon \gt 0\), there is some \(N\) such that for all \(n \geqslant N\), \[ \left| \frac{2n - 3}{3n + 7} - \frac{2}{3} \right| \lt \varepsilon. \]
    Manipulating the Expression:
    We start by simplifying the difference between \(x_n = \frac{2n - 3}{3n + 7}\) and \(\frac{2}{3}\): \[ \left| \frac{2n - 3}{3n + 7} - \frac{2}{3} \right|. \] To combine these two terms into a single fraction, we first find a common denominator: \[ \frac{2n - 3}{3n + 7} - \frac{2}{3} = \frac{(2n - 3)\cdot 3 - 2(3n + 7)}{3(3n + 7)}. \] Simplifying the numerator: \[ (2n - 3)\cdot 3 = 6n - 9, \quad 2(3n + 7) = 6n + 14, \] so the numerator becomes: \[ (6n - 9) - (6n + 14) = 6n - 9 - 6n - 14 = -23. \] Thus, we have: \[ \frac{2n - 3}{3n + 7} - \frac{2}{3} = \frac{-23}{3(3n + 7)}. \]
    Bounding the Expression:
    Now, we need to show that this expression becomes small as \(n\) increases. Specifically, we want to show that for any \(\varepsilon \gt 0\), there exists \(N\) such that for all \(n \geqslant N\), \[ \left| \frac{-23}{3(3n + 7)} \right| \lt \varepsilon. \] Simplifying further: \[ \left| \frac{-23}{3(3n + 7)} \right| = \frac{23}{3(3n + 7)}. \] Since \(3n + 7\) grows without bound as \(n\) increases, this fraction tends to zero. To make it less than \(\varepsilon\), we solve the inequality: \[ \frac{23}{3(3n + 7)} \lt \varepsilon. \] Multiplying both sides by \(3(3n + 7)\) and solving for \(n\): \[ 23 \lt 3\varepsilon (3n + 7), \] \[ \frac{23}{3\varepsilon} \lt 3n + 7, \] \[ \frac{23}{3\varepsilon} - 7 \lt 3n, \] \[ n \gt \frac{23}{9\varepsilon} - \frac{7}{3}. \] Let \(N = \left\lceil \frac{23}{9\varepsilon} - \frac{7}{3} \right\rceil\). (Note: \(\lceil x\rceil\) est la valeur entière supérieure de \(x\), c'est-à-dire le plus petit entier \(n\geqslant x\).) For all \(n \geqslant N\), we have: \[ \left| \frac{-23}{3(3n + 7)} \right| \lt \varepsilon. \]
    Conclusion:
    We have shown that for any \(\varepsilon \gt 0\), there exists \(N\) such that for all \(n \geqslant N\), \[ \left| x_n - \frac{2}{3} \right| \lt \varepsilon, \] which means that: \[ \lim_{n \to \infty} x_n = \frac{2}{3}. \]

  2. En préparation, étudions la différence \[\begin{aligned} |x_n-\tfrac{a}{c}|&= \Bigl| \frac{an+b}{cn+d}-\frac{a}{c} \Bigr|\\ &= \Bigl| \frac{c(an+b)-a(cn+d)}{c(cn+d)} \Bigr|\\ &= \frac{|ad-bc|}{|c(cn+d)|} = \frac{|ad-bc|}{c} \frac{1}{cn+d} \end{aligned}\] Dans la dernière égalité, on a utilisé le fait que \(|cn+d|=cn+d\) (car \(c\gt 0\), \(d\gt 0\) et \(n\geqslant 1\)). Si maintenant on fixe \(\varepsilon\gt 0\), on aura \(|x_n-\frac{a}{c}|\leqslant \varepsilon\) si et seulement si \[ \frac{|ad-bc|}{c} \frac{1}{cn+d}\leqslant \varepsilon \quad\Leftrightarrow\quad n\geqslant \frac{1}{c}\left( \frac{|ad-bc|}{c\varepsilon}-d \right)\,. \] Ceci signifie qu'il existe \(N\) tel que \(|x_n-\frac{a}{c}|\leqslant \varepsilon\) pour tout \(n\geqslant N\).