{"id":229649,"date":"2025-06-08T10:43:09","date_gmt":"2025-06-08T10:43:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=229649"},"modified":"2025-06-08T10:43:11","modified_gmt":"2025-06-08T10:43:11","slug":"finding-limits-from-graphs-calculus-ca1-name-1","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/08\/finding-limits-from-graphs-calculus-ca1-name-1\/","title":{"rendered":"Finding Limits from Graphs Calculus CA#1 Name: 1"},"content":{"rendered":"\n

Finding Limits from Graphs Calculus CA#1 Name: 1. a. lim f(x) = x g. lim f(x) = b. f(-3) = c. lim f(x) = e. f(2) = f. lim f(x) = x^2 h. f(4) = i. lim f(x) = 2. a. lim f(x) = b. f(-2) = c. lim f(x) = d. lim f(x) = e. f(-4) = f. lim f(x) = g. lim f(x) = h. f(-5) = i. f(1) = 3. Sketch a graph of a function f that satisfies all of the following conditions. a. f(3) = 4 b. lim f(x) = 2 c. lim f(x) = -4 d. f(-2) is undefined. e. lim f(x) > lim f(x)<\/p>\n\n\n\n

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

Sample Answers (Based on a Generic Piecewise Graph)<\/strong><\/h3>\n\n\n\n

1.<\/strong><\/h4>\n\n\n\n

a. lim\u2061x\u2192\u22123f(x)=x\\lim_{x \\to -3} f(x) = xlimx\u2192\u22123\u200bf(x)=x
b. f(\u22123)=(value from the graph)f(-3) = \\text{(value from the graph)}f(\u22123)=(value from the graph)
c. lim\u2061x\u21922f(x)=(approaching value from both sides)\\lim_{x \\to 2} f(x) = \\text{(approaching value from both sides)}limx\u21922\u200bf(x)=(approaching value from both sides)
e. f(2)=(value at x=2)f(2) = \\text{(value at } x = 2)f(2)=(value at x=2)
f. lim\u2061x\u2192x2f(x)=(nonsensical unless \u2019x\u2019 is replaced with a number)\\lim_{x \\to x^2} f(x) = \\text{(nonsensical unless ‘x’ is replaced with a number)}limx\u2192x2\u200bf(x)=(nonsensical unless \u2019x\u2019 is replaced with a number)
g. lim\u2061x\u21924f(x)=b\\lim_{x \\to 4} f(x) = blimx\u21924\u200bf(x)=b
h. f(4)=(value at x=4)f(4) = \\text{(value at } x = 4)f(4)=(value at x=4)
i. lim\u2061x\u21921f(x)=(value approached near 1)\\lim_{x \\to 1} f(x) = \\text{(value approached near 1)}limx\u21921\u200bf(x)=(value approached near 1)<\/p>\n\n\n\n

2.<\/strong><\/h4>\n\n\n\n

a. lim\u2061x\u2192\u22122f(x)=(left = right?)\\lim_{x \\to -2} f(x) = \\text{(left = right?)}limx\u2192\u22122\u200bf(x)=(left = right?)
b. f(\u22122)=actual value at x=\u22122f(-2) = \\text{actual value at } x = -2f(\u22122)=actual value at x=\u22122
c. lim\u2061x\u21920f(x)=\u2026\\lim_{x \\to 0} f(x) = \\ldotslimx\u21920\u200bf(x)=\u2026
d. lim\u2061x\u2192\u22121f(x)=\u2026\\lim_{x \\to -1} f(x) = \\ldotslimx\u2192\u22121\u200bf(x)=\u2026
e. f(\u22124)=\u2026f(-4) = \\ldotsf(\u22124)=\u2026
f. lim\u2061x\u21921f(x)=\u2026\\lim_{x \\to 1} f(x) = \\ldotslimx\u21921\u200bf(x)=\u2026
g. lim\u2061x\u2192\u22125f(x)=\u2026\\lim_{x \\to -5} f(x) = \\ldotslimx\u2192\u22125\u200bf(x)=\u2026
h. f(\u22125)=\u2026f(-5) = \\ldotsf(\u22125)=\u2026
i. f(1)=\u2026f(1) = \\ldotsf(1)=\u2026<\/p>\n\n\n\n

\n\n\n\n

3. Construct a Graph with These Conditions:<\/strong><\/h3>\n\n\n\n

a. f(3)=4f(3) = 4f(3)=4: So, there’s a point at (3, 4)<\/strong>.
b. lim\u2061x\u21923f(x)=2\\lim_{x \\to 3} f(x) = 2limx\u21923\u200bf(x)=2: The graph approaches 2 from both sides<\/strong>, but there is a hole at x=3x = 3x=3.
c. lim\u2061x\u2192\u22122f(x)=\u22124\\lim_{x \\to -2} f(x) = -4limx\u2192\u22122\u200bf(x)=\u22124: So the curve gets close to \u22124-4\u22124 from both sides as x\u2192\u22122x \\to -2x\u2192\u22122.
d. f(\u22122)f(-2)f(\u22122) is undefined: No point at x=\u22122x = -2x=\u22122.
e. lim\u2061x\u21920\u2212f(x)>lim\u2061x\u21920+f(x)\\lim_{x \\to 0^-} f(x) > \\lim_{x \\to 0^+} f(x)limx\u21920\u2212\u200bf(x)>limx\u21920+\u200bf(x): A jump discontinuity<\/strong> at 0.<\/p>\n\n\n\n

\n\n\n\n

Explanation<\/strong><\/h3>\n\n\n\n

Finding limits from graphs is a fundamental concept in calculus that helps us understand the behavior of a function near a specific point. The limit<\/strong> of a function f(x)f(x)f(x) as xxx approaches a value aaa, written lim\u2061x\u2192af(x)\\lim_{x \\to a} f(x)limx\u2192a\u200bf(x), describes the value that the function approaches<\/strong>, not necessarily the value at x=ax = ax=a.<\/p>\n\n\n\n

To evaluate limits from a graph, observe the left-hand limit<\/strong> and right-hand limit<\/strong>. If both sides approach the same value, the limit exists. If they approach different values, the limit does not exist (DNE)<\/strong>. For example, if lim\u2061x\u21923\u2212f(x)=2\\lim_{x \\to 3^-} f(x) = 2limx\u21923\u2212\u200bf(x)=2 and lim\u2061x\u21923+f(x)=2\\lim_{x \\to 3^+} f(x) = 2limx\u21923+\u200bf(x)=2, then lim\u2061x\u21923f(x)=2\\lim_{x \\to 3} f(x) = 2limx\u21923\u200bf(x)=2. However, the function value f(3)f(3)f(3) could be different or even undefined, which shows the importance of distinguishing between a function’s limit and its actual value.<\/p>\n\n\n\n

Discontinuities are key features to recognize. Removable discontinuities<\/strong> occur when the limit exists but the function value is missing or different (a hole in the graph). Jump discontinuities<\/strong> occur when the left-hand and right-hand limits differ. Infinite discontinuities<\/strong> occur when the function goes to infinity.<\/p>\n\n\n\n

In problem 3, we see all these ideas in play: the function approaches a value but is defined differently at the point (like at x=3x = 3x=3), or the function isn\u2019t defined at all (like at x=\u22122x = -2x=\u22122). Constructing a graph with these constraints helps reinforce understanding of how limits behave and how function values may differ from them.<\/p>\n\n\n\n

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

Finding Limits from Graphs Calculus CA#1 Name: 1. a. lim f(x) = x g. lim f(x) = b. f(-3) = c. lim f(x) = e. f(2) = f. lim f(x) = x^2 h. f(4) = i. lim f(x) = 2. a. lim f(x) = b. f(-2) = c. lim f(x) = d. lim f(x) = […]<\/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-229649","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229649","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=229649"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/229649\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=229649"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=229649"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=229649"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}