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Calculus I

by: Anibal Schmidt

Calculus I MATH 1441

Anibal Schmidt
GPA 3.52

Marshall Ransom

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Marshall Ransom
Class Notes
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This 2 page Class Notes was uploaded by Anibal Schmidt on Monday October 12, 2015. The Class Notes belongs to MATH 1441 at Georgia Southern University taught by Marshall Ransom in Fall. Since its upload, it has received 11 views. For similar materials see /class/222042/math-1441-georgia-southern-university in Mathematics (M) at Georgia Southern University.

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Date Created: 10/12/15
MATH 1441 Spring 2006 NAME Instructor Mr Ransom The limit of a function from left or right or both If x 3x2 x 7 2 we call the limit of this function as x approaches 1 from the left as l 312 1 7 2 2 Similarly we call the limit as x approaches 1 from the right 2 What is mysterious about this concept of limit The function obviously has a value of 2 when x 1 We say that lin ll fx 1in3xl x 2 2 This is a simple thing to analyze because this function is X X continuous the graph is connected and we can clearly see that the function value at x 1 is 2 See the graph below IHDUM 39 Xm1n395 Hmax5 H5911 Vm1n39 1 2 S th t 2x3 if xgt 1 u ose a x pp g 4 x2 if xS l We get two different values for this limit because there are two different expressions for x Ifx approaches 7 1 from the left we use the formula 4 7 x2 which gives us 4 7 7 12 Ifx approaches 7 1 from the right we use the formula 72x 3 which gives us 7 27 1 3 5 What is the limit as x approaches 7 1 from the leftright In this case we say that xgn gx 3 and xgn gx 5 Since these limits are different we say that the ONE limit asx approaches 71 does not exist In order for this limit to exist both the left hand and right hand limits would have to be the same and the graph would have to connec from the left and right sides These limits from the left and right have different values Looking at a graph from a calculator screen we can see that the left hand graph and the right hand graph do not meet in one point but the limits from the left and right sides can be seen on the graph as the y values of this lnction for each piecewisedefined part of the graph Where should there be an open circle on the graph 1 3 Graph of gx E If I H 1 Notice that the value of the function gx at the point 7 1 is g71 4 7 7 12 3 because 4 7 x2 de nes this lnction for x 71 This is an important fact as we examine the continuity of a function We will compare this value if it exists to the limit value F915 Continuity of a function at a point x a KNOW THIS DEFINITION l A lnction is said to be continuous at a point where x a if three conditions hold i faexists ii fx exists iii f x f a 2x3 if xgt l Usmg our functlon gx we apply the de mtlon of cont1nu1ty at the p01nt x fl 4 x2 1f x S 1 Condition i is OK because g 1 3 meaning that g 1 exists Condition ii is not OK because lin gx 3 7E lin gx 5 meaning lim1gx does not exist xgtr x xgtr We STOP here because as soon as one of these three conditions goes wrong we know the function is NOT continuous at the given value of x Try these Find the following limits if they exist 1 139 35 2139 f39ff 3mm 3139 f 3mm 1mx 1mx1x 1mx1x 90 xvi xlif xlt2 xvzg g x8ifxlt2 Use x and gx from problems 2 and 3 above 4 lim fx 5 lim fx 6 lim gx 7 lim gx xgt2 x xgt2 x x2 8 lngc 9 lggbc 10 x2 ll x3 12 x2 13 Discuss the continuity of x and gx at x 2 2 x2 9 x 9 1f x at 3 and rx x 3 and d1scuss cont1nu1ty at x 3 4 if x3 l4 Graph hx x


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