{"id":251659,"date":"2025-07-10T19:18:20","date_gmt":"2025-07-10T19:18:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=251659"},"modified":"2025-07-10T19:18:22","modified_gmt":"2025-07-10T19:18:22","slug":"to-appreciate-the-dirac-delta-function-as-the-limit-of-a-certain-function-that-is-more-elementary-consider-x-i%c2%b5-a%e2%80%a0-gi%c2%b5x-8x-i%c2%b5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/to-appreciate-the-dirac-delta-function-as-the-limit-of-a-certain-function-that-is-more-elementary-consider-x-i%c2%b5-a%e2%80%a0-gi%c2%b5x-8x-i%c2%b5\/","title":{"rendered":"To appreciate the Dirac delta function as the limit of a certain function that is more elementary, consider |x| < \u00ce\u00b5 \u00e2\u2020\u2019 g\u00ce\u00b5(x) = 8|x| > \u00ce\u00b5"},"content":{"rendered":"\n

To appreciate the Dirac delta function as the limit of a certain function that is more elementary, consider |x| < \u00ce\u00b5 \u00e2\u2020\u2019 g\u00ce\u00b5(x) = 8|x| > \u00ce\u00b5. Plot the functions g\u00ce\u00b5(x) and g(x0) for \u00ce\u00b5 = 0.1, 0.01, and \u00ce\u00b5 \u00e2\u2020\u2019 0. Show that lim g\u00ce\u00b5(x) = 0 for \u00ce\u00b5 \u00e2\u2020\u2019 0. Show that lim g\u00ce\u00b5(0) = 0 for \u00ce\u00b5 \u00e2\u2020\u2019 0. Show that \u00e2\u02c6\u00abg\u00ce\u00b5(\u00ce\u00b50) dx = 1 for any \u00ce\u00b50 and \u00e2\u02c6\u00abf(c)g(x0) dx = f(\u00ce\u00b50) for any \u00ce\u00b50 where f is continuous.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The Dirac delta function is often considered in the context of distributions, which are not functions in the traditional sense but rather generalized functions that act on test functions. A standard approach to defining the Dirac delta function, \u03b4(x)\\delta(x)\u03b4(x), is through its limiting behavior as a sequence of functions.<\/p>\n\n\n\n

The function g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x):<\/h3>\n\n\n\n

Given that you want to appreciate the Dirac delta function as the limit of a simpler function, consider g\u03f5(x)=8\u2223x\u2223\u03f5g_\\epsilon(x) = \\frac{8|x|}{\\epsilon}g\u03f5\u200b(x)=\u03f58\u2223x\u2223\u200b for \u2223x\u2223<\u03f5|x| < \\epsilon\u2223x\u2223<\u03f5 and g\u03f5(x)=0g_\\epsilon(x) = 0g\u03f5\u200b(x)=0 otherwise.<\/p>\n\n\n\n

This function is a piecewise linear function that has a “spike” at x=0x = 0x=0, with a peak that becomes increasingly narrow as \u03f5\u21920\\epsilon \\to 0\u03f5\u21920, while its total area remains constant.<\/p>\n\n\n\n

    \n
  1. Plotting g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x)<\/strong> for different values of \u03f5\\epsilon\u03f5:\n
      \n
    • For \u03f5=0.1\\epsilon = 0.1\u03f5=0.1, g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) will be a triangle with base 2\u03f5=0.22\\epsilon = 0.22\u03f5=0.2 and height 808080 at x=0x = 0x=0.<\/li>\n\n\n\n
    • For \u03f5=0.01\\epsilon = 0.01\u03f5=0.01, the triangle becomes much narrower and taller, with a height of 800800800 at x=0x = 0x=0.<\/li>\n\n\n\n
    • As \u03f5\u21920\\epsilon \\to 0\u03f5\u21920, the function becomes increasingly concentrated around x=0x = 0x=0, approaching the Dirac delta function.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Limit as \u03f5\u21920\\epsilon \\to 0\u03f5\u21920<\/strong>:\n
        \n
      • For \u03f5\u21920\\epsilon \\to 0\u03f5\u21920, the function g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) tends to zero for any x\u22600x \\neq 0x\ue020=0 because the area under the function remains constant while the base of the triangle narrows.<\/li>\n\n\n\n
      • Therefore, lim\u2061\u03f5\u21920g\u03f5(x)=0\\lim_{\\epsilon \\to 0} g_\\epsilon(x) = 0lim\u03f5\u21920\u200bg\u03f5\u200b(x)=0 for all x\u22600x \\neq 0x\ue020=0.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Evaluating at x=0x = 0x=0<\/strong>:\n
          \n
        • At x=0x = 0x=0, g\u03f5(0)=8\u22230\u2223\u03f5=0g_\\epsilon(0) = \\frac{8|0|}{\\epsilon} = 0g\u03f5\u200b(0)=\u03f58\u22230\u2223\u200b=0 for all \u03f5\\epsilon\u03f5. Hence, lim\u2061\u03f5\u21920g\u03f5(0)=0\\lim_{\\epsilon \\to 0} g_\\epsilon(0) = 0lim\u03f5\u21920\u200bg\u03f5\u200b(0)=0.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Integral of g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x)<\/strong>:\n
            \n
          • The integral of g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) over R\\mathbb{R}R is given by: \u222b\u2212\u221e\u221eg\u03f5(x)\u2009dx=\u222b\u2212\u03f5\u03f58\u2223x\u2223\u03f5\u2009dx=1.\\int_{-\\infty}^{\\infty} g_\\epsilon(x) \\, dx = \\int_{-\\epsilon}^{\\epsilon} \\frac{8|x|}{\\epsilon} \\, dx = 1.\u222b\u2212\u221e\u221e\u200bg\u03f5\u200b(x)dx=\u222b\u2212\u03f5\u03f5\u200b\u03f58\u2223x\u2223\u200bdx=1. The area under the triangle is always 1, regardless of the value of \u03f5\\epsilon\u03f5.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Action of g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) on a continuous function<\/strong>:\n
              \n
            • For any continuous function f(x)f(x)f(x), the action of g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) is given by: \u222b\u2212\u221e\u221ef(x)g\u03f5(x)\u2009dx=f(0).\\int_{-\\infty}^{\\infty} f(x) g_\\epsilon(x) \\, dx = f(0).\u222b\u2212\u221e\u221e\u200bf(x)g\u03f5\u200b(x)dx=f(0). This is because the function g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) is nonzero only in a narrow region around x=0x = 0x=0, and as \u03f5\\epsilon\u03f5 becomes smaller, the integral becomes more concentrated at x=0x = 0x=0, thus approaching f(0)f(0)f(0).<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

              Conclusion:<\/h3>\n\n\n\n

              The function g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) approaches the Dirac delta function \u03b4(x)\\delta(x)\u03b4(x) as \u03f5\u21920\\epsilon \\to 0\u03f5\u21920, because:<\/p>\n\n\n\n

                \n
              • lim\u2061\u03f5\u21920g\u03f5(x)=0\\lim_{\\epsilon \\to 0} g_\\epsilon(x) = 0lim\u03f5\u21920\u200bg\u03f5\u200b(x)=0 for x\u22600x \\neq 0x\ue020=0,<\/li>\n\n\n\n
              • g\u03f5(0)=0g_\\epsilon(0) = 0g\u03f5\u200b(0)=0 for all \u03f5\\epsilon\u03f5,<\/li>\n\n\n\n
              • \u222b\u2212\u221e\u221eg\u03f5(x)\u2009dx=1\\int_{-\\infty}^{\\infty} g_\\epsilon(x) \\, dx = 1\u222b\u2212\u221e\u221e\u200bg\u03f5\u200b(x)dx=1 for any \u03f5\\epsilon\u03f5,<\/li>\n\n\n\n
              • \u222b\u2212\u221e\u221ef(x)g\u03f5(x)\u2009dx\\int_{-\\infty}^{\\infty} f(x) g_\\epsilon(x) \\, dx\u222b\u2212\u221e\u221e\u200bf(x)g\u03f5\u200b(x)dx approaches f(0)f(0)f(0) for continuous functions f(x)f(x)f(x).<\/li>\n<\/ul>\n\n\n\n

                This demonstrates that the Dirac delta function is the limiting form of g\u03f5(x)g_\\epsilon(x)g\u03f5\u200b(x) as \u03f5\u21920\\epsilon \\to 0\u03f5\u21920.<\/p>\n\n\n\n

                \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                To appreciate the Dirac delta function as the limit of a certain function that is more elementary, consider |x| < \u00ce\u00b5 \u00e2\u2020\u2019 g\u00ce\u00b5(x) = 8|x| > \u00ce\u00b5. Plot the functions g\u00ce\u00b5(x) and g(x0) for \u00ce\u00b5 = 0.1, 0.01, and \u00ce\u00b5 \u00e2\u2020\u2019 0. Show that lim g\u00ce\u00b5(x) = 0 for \u00ce\u00b5 \u00e2\u2020\u2019 0. Show that lim […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-251659","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251659","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=251659"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/251659\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=251659"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=251659"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=251659"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}