{"id":27952,"date":"2024-08-16T13:00:42","date_gmt":"2024-08-16T13:00:42","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=27952"},"modified":"2024-08-15T22:38:58","modified_gmt":"2024-08-15T22:38:58","slug":"le-theoreme-du-sandwich-pour-le-calcul-des-limites","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/le-theoreme-du-sandwich-pour-le-calcul-des-limites\/","title":{"rendered":"Le Th\u00e9or\u00e8me du Sandwich pour le Calcul des Limites"},"content":{"rendered":"<p><center><\/p>\n<h1>Le Th\u00e9or\u00e8me du Sandwich pour le Calcul des Limites<\/h1>\n<p><em><strong>R\u00e9sum\u00e9:<\/strong><br \/>\nCe cours pr\u00e9sente le Th\u00e9or\u00e8me du Sandwich, un outil cl\u00e9 en calcul pour \u00e9valuer des limites difficiles en utilisant des fonctions plus simples qui les encadrent par le haut et par le bas. Une explication graphique et une d\u00e9monstration formelle sont propos\u00e9es, suivies d&#8217;exemples pratiques. L&#8217;objectif est que les \u00e9tudiants comprennent comment appliquer ce th\u00e9or\u00e8me pour calculer les limites de mani\u00e8re plus efficace.<\/em><\/p>\n<p><strong>Objectifs d&#8217;apprentissage:<\/strong><br \/>\n\u00c0 la fin de ce cours, l&#8217;\u00e9tudiant sera capable de<\/p>\n<ul style=\"text-align:left;\">\n<li><strong>Comprendre<\/strong> l&#8217;utilit\u00e9 du Th\u00e9or\u00e8me du Sandwich dans le calcul des limites.<\/li>\n<li><strong>Identifier<\/strong> des fonctions qui peuvent encadrer une fonction cible pour appliquer le th\u00e9or\u00e8me.<\/li>\n<li><strong>Appliquer<\/strong> le Th\u00e9or\u00e8me du Sandwich pour calculer des limites difficiles.<\/li>\n<li><strong>Visualiser<\/strong> graphiquement le concept du Th\u00e9or\u00e8me du Sandwich.<\/li>\n<li><strong>D\u00e9montrer<\/strong> le Th\u00e9or\u00e8me du Sandwich de mani\u00e8re formelle.<\/li>\n<\/ul>\n<p><strong><u>TABLE DES MATI\u00c8RES<\/u>:<\/strong><br \/>\n<a href=\"#1\">Introduction<\/a><br \/>\n<a href=\"#2\">Id\u00e9e Graphique du Th\u00e9or\u00e8me du Sandwich<\/a><br \/>\n<a href=\"#3\">D\u00e9monstration du Th\u00e9or\u00e8me du Sandwich<\/a><br \/>\n<a href=\"#4\">Exemples<\/a><\/p>\n<p><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/24G_qlEwL9M\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><br \/>\n<\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>Introduction<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=24G_qlEwL9M&amp;t=158s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">L&#8217;utilit\u00e9 du Th\u00e9or\u00e8me du Sandwich r\u00e9side dans la facilit\u00e9 qu&#8217;il offre pour calculer certaines limites difficiles<\/span><\/strong><\/a> \u00e0 travers d&#8217;autres plus simples. Le nom vient du fait que, au lieu de calculer directement la limite d&#8217;une fonction lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x\\to x_0<\/span><\/span>, on utilise une autre paire de fonctions, l&#8217;une encadrant par le haut et l&#8217;autre par le bas, et dont la limite en <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0<\/span><\/span> co\u00efncide et est facile \u00e0 obtenir. Comme la fonction d&#8217;origine est toujours entre les deux, elle est comme \u00able fromage entre deux tranches de pain\u00bb.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Id\u00e9e Graphique du Th\u00e9or\u00e8me du Sandwich<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=24G_qlEwL9M&amp;t=206s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">L&#8217;id\u00e9e qui synth\u00e9tise le th\u00e9or\u00e8me est en r\u00e9alit\u00e9 assez simple.<\/span> <\/strong><\/a>Supposons que nous voulons calculer une limite difficile<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle\\lim_{x \\to x_0}f(x)<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Ce que l&#8217;on fait g\u00e9n\u00e9ralement, c&#8217;est de prendre toutes nos connaissances en alg\u00e8bre des fonctions pour essayer <strong>de la simplifier au point o\u00f9 nous pouvons l&#8217;\u00e9valuer<\/strong>. Cependant, parfois, une approche diff\u00e9rente est beaucoup plus efficace. Supposons que nous avons un intervalle ferm\u00e9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">I<\/span><\/span> tel que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0 \\in I<\/span><\/span> et qu&#8217;il existe deux autres fonctions <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">m(x)<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">M(x)<\/span><\/span> qui satisfont la relation suivante<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall x\\in I)(m(x)\\leq f(x) \\leq M(x) )<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Et qu&#8217;en plus<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to x_0} m(x) = \\lim_{x\\to x_0} M(x) = L<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Alors, il se v\u00e9rifiera que<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to x_0} f(x) = L<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">C&#8217;est ce que nous pouvons voir dans l&#8217;image suivante.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-gjBfVdaLj-k\/YGDXVUQBqDI\/AAAAAAAAEvg\/d2sNJdweVaoB64O5e2qfBxjZGIyIyGOxgCLcBGAsYHQ\/s0\/teorema%2Bdel%2Bsandwich.PNG\" alt=\"th\u00e9or\u00e8me du sandwich\" class=\"alignnone size-full lazyload\" width=\"513\" height=\"407\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-gjBfVdaLj-k\/YGDXVUQBqDI\/AAAAAAAAEvg\/d2sNJdweVaoB64O5e2qfBxjZGIyIyGOxgCLcBGAsYHQ\/s0\/teorema%2Bdel%2Bsandwich.PNG\" alt=\"th\u00e9or\u00e8me du sandwich\" class=\"alignnone size-full lazyload\" width=\"513\" height=\"407\" \/><\/noscript><\/center><\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>D\u00e9monstration du Th\u00e9or\u00e8me du Sandwich<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=24G_qlEwL9M&amp;t=404s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Pour d\u00e9montrer le Th\u00e9or\u00e8me du Sandwich<\/span><\/strong><\/a>, nous suivrons le raisonnement suivant :<\/p>\n<table style=\"text-align: justify; color: #000000;\">\n<tbody>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1)<\/span><\/span><\/td>\n<td><span style=\"background-color: #90ff90;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x_0\\in I<\/span><\/span><\/span>; <strong>Pr\u00e9misse<\/strong><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2)<\/span><\/span><\/td>\n<td><span style=\"background-color: #90ff90;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to x_0} m(x) = L<\/span><\/span><\/span> ; <strong>Pr\u00e9misse<\/strong><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\epsilon \\gt 0)(\\exists \\delta_1 \\gt 0) (|x-x_0|\\lt \\delta_1 \\rightarrow |m(x) -L| \\lt \\epsilon )<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(3)<\/span><\/span><\/td>\n<td><span style=\"background-color: #90ff90;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to x_0} M(x) = L<\/span><\/span><\/span> ; <strong>Pr\u00e9misse<\/strong><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\epsilon \\gt 0)(\\exists \\delta_2 \\gt 0) (|x-x_0|\\lt \\delta_2 \\rightarrow |M(x) -L| \\lt \\epsilon )<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(4)<\/span><\/span><\/td>\n<td><span style=\"background-color: #90ff90;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall x \\in I)(m(x) \\leq f(x) \\leq M(x) )<\/span><\/span><\/span>; <strong>Pr\u00e9misse<\/strong><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(5)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall x \\in I)(m(x) - L \\leq f(x) - L \\leq M(x) - L )<\/span><\/span>; De (4)<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(6)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(|m(x) -L|\\lt \\epsilon) \\rightarrow (-\\epsilon \\lt m(x) - L \\lt \\epsilon)<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(7)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(|M(x) -L|\\lt \\epsilon ) \\rightarrow (-\\epsilon \\lt M(x) - L \\lt \\epsilon) <\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(8)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\epsilon \\gt 0)(\\exists \\delta \\gt 0) (|x-x_0|\\lt \\delta=\\min\\{\\delta_1,\\delta_2\\} \\rightarrow ( |M(x) -L| \\lt \\epsilon \\wedge |m(x) -L| \\lt \\epsilon ) )<\/span><\/span>; De (2,3)<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(9)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\epsilon \\gt 0)(\\exists \\delta \\gt 0) (|x-x_0|\\lt \\delta=\\min\\{\\delta_1,\\delta_2\\} \\rightarrow ( - \\epsilon \\lt f(x) - L \\lt \\epsilon ) )<\/span><\/span>; De (1,5,6,7,8)<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\epsilon \\gt 0)(\\exists \\delta \\gt 0) (|x-x_0|\\lt \\delta=\\min\\{\\delta_1,\\delta_2\\} \\rightarrow |f(x) - L| \\lt \\epsilon ) )<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to x_0}f(x) = L\\;\\blacksquare<\/span><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Exemples<\/h2>\n<p style=\"text-align: justify; color: #000000;\">En utilisant le Th\u00e9or\u00e8me du Sandwich, nous pouvons calculer la limite de fonctions m\u00eame lorsque nous n&#8217;avons pas leur expression alg\u00e9brique explicite. Un exemple de cela se pr\u00e9sente dans la situation suivante :<\/p>\n<p style=\"text-align: justify; color: #000000;\">Un exemple de cela se pr\u00e9sente dans la situation suivante :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li>Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sqrt{5-2x^2}\\leq f(x) \\leq \\sqrt{5-x^2}<\/span><\/span>, lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">-1\\leq x\\leq 1<\/span><\/span>. Quelle est la valeur de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to 0}f(x)<\/span><\/span>? <a href=\"https:\/\/www.youtube.com\/watch?v=24G_qlEwL9M&amp;t=1082s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">[SOLUTION]<\/span><\/strong><\/a><\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">Une autre utilisation pratique du Th\u00e9or\u00e8me du Sandwich se produit lorsque la limite elle-m\u00eame n&#8217;est pas \u00e9vidente par rapport \u00e0 d&#8217;autres limites plus simples qui l&#8217;encadrent par le haut et par le bas, comme c&#8217;est le cas lorsque l&#8217;on calcule l&#8217;exemple suivant :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li>Calculer : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\lim_{x\\to 0}\\dfrac{\\sin(x)}{x}<\/span><\/span> <a href=\"https:\/\/www.youtube.com\/watch?v=24G_qlEwL9M&amp;t=1157s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">[SOLUTION]<\/span><\/strong><\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Le Th\u00e9or\u00e8me du Sandwich pour le Calcul des Limites R\u00e9sum\u00e9: Ce cours pr\u00e9sente le Th\u00e9or\u00e8me du Sandwich, un outil cl\u00e9 en calcul pour \u00e9valuer des limites difficiles en utilisant des fonctions plus simples qui les encadrent par le haut et par le bas. Une explication graphique et une d\u00e9monstration formelle sont propos\u00e9es, suivies d&#8217;exemples pratiques. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":27930,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":188,"footnotes":""},"categories":[866,569],"tags":[],"class_list":["post-27952","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calcul-differentiel","category-mathematiques"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Le Th\u00e9or\u00e8me du Sandwich pour le Calcul des Limites - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Apprenez \u00e0 calculer des limites en utilisant le Th\u00e9or\u00e8me du Sandwich. 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