51707-772886

PROGRAM: BACHELOR OF RISK MANAGEMENT AND TAKAFUL

FIRST YEAR ( SEMESTER 1) 2014/2015

COURSE: BUSINESS MATHEMATICS ( MQF 3013)

TOPIC 6: EVALUATING LIMIT

– USING TABLE AND GRAPH

– USING ALGEBRA METHOD

Submitted To :

Dr. Puspa Liza Binti Ghazali

Submitted By:

GROUP MEMBER’S NAME NUMBER MATRIC

Siti Hajar Binti Mustafa 038307

Siti Nur fatin Binti Abdul Fatah 037300

Wan Aisyah Munirah Binti Wan Lokman 037406

Nur Aqilah Binti Adnan 037093

Norazietul Wahida Binti Ramadzan 037672

INTRODUCTION

The first part of the tutorial contains a list of concept that be used to evaluate many limits. The second part contains a collection of examples that these theorems cannot be used to evaluate immediately. It is shown how to do some algebraic manipulation to put these examples in the form so that the concept can be applied. After working through these materials, the student should know this basic concept, how to apply them to evaluate limits and how to manipulate certain examples so that the concept may be used. Objectives modules used in this chapter is to help us in the symbolic or algebraic computation of limits, we have a list of limit concept. As a result of this concept, we see that for many functions f(x). One of the functions which has this property is called continuous. From the above-mentioned list of limit concept, we see that polynomial functions and rational functions are continuous. We will study continuous functions more extensively in another module. The following examples demonstrate how we can evaluate limits of functions which are not continuous by using the above-mentioned list of limit concepts. These include many of the examples which were explored numerically and/or graphically.

CONCEPT OF LIMITS

Limit is the value that a function or sequence “approaches” as the input or index approaches some value. Limits are essential to calculus (and mathematical analysis in general) and are used to define continuity, derivatives, and integrals.

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If the function f approaches a number L when x approaches a, then we say that the limit of f(x) when x approaches a is L and is written as limx?af(x) =L.

Basic properties of limits.

limx?ac=climx?axn =anlimx?afx± gx =limx?afx± limx?a gx limx?afx. gx =limx?afx . limx?agx limx?a f(x)g(x) = limx?a f(x)limx?a g(x) , (limx?a gx?0)limx?a c fx=c limx?a f(x) If limx?a f(x)g(x) = 00 , which is an indeterminate form, then the functions f(x) and gxneed to be simplified by factorization or by multiplying with their conjugates.

Example: limx?1×2-1x-1By factoring (x2?1) into (x?1)(x+1) we get,

limx?1×2-1x-1= limx?1x-1x+1x-1= limx?1x+1Now we can just substitute x=1 to get the limit,

limx?1(x+1) = 1+1 = 2

If limx?? f(x)g(x) = ?? , which is an indeterminate form, then we need to simplify the function by dividing the numerator and the denominator by the highest power of x of the denominator.(or simply by the highest power of x)

Example :

Since is the highest power of x in the denominator of our function, we have

Example for graphing method :

limx?1×3-x2+3x-3x-1 ==

=

4

Table :

X 0 0.5 0.7 0.8 0.9 0.99 0.999

f(x) 3 3.25 3.49 3.64 3.81 3.98 3.998

-123825-400050Example 1: (Siti Hajar Bt Mustafa)

00Example 1: (Siti Hajar Bt Mustafa)

Let the equation given is,limx?2(x2-1)(x-1)The way to know the limit of x is,limx?2(x2-1)x-1Let work it (x = 1).So, it will be limx?2(12-1)(1-1) =00It was show that value of x cannot be 1, except below or upper than 1. After knew the value of x now to get the value of f(x) based on number of x.

x 0.5 0.9 0.99 0.999 0.9999 0.99999

f(x) 1.5 1.9 1.99 1.999 1.9999 1.99999

Substitute the value of x into the equation given: Example,

279654020574000f(x)=limx?2×2- 1(x-1)Equation Given

27965401739900028295604318000300418524066500f(x)=limx?2x-1x+1(x-1) Simplify

(x=0.5) = 0.5 + 1 Substitute

f(x)=1.5

Need to use the same way to get all the value of f(x).After get the all value, Need to draw the graph based on the both value.

Example 2 : (Norazietul Wahida Binti Ramadzan)

282765515049500limx?3 2×2-8x+6x-3By factoring into (x-3) (x-1) we get:

282765517716500limx?3 2×2-8x+6x-3291211013208000=limx?3x-3x-1(x-3)=limx?3(x-1) So, the limit of (2x²-8x+6) as x approaches 3 is 0

x-3

limx?3 x-3=3-3=0x 1 2 3

F(x) -2 -1 0

Now, we can just substitute x=3 to get the limit

Example 3 : (Siti Nur Fatin Binti Abdul Fatah)

203327049212500224218528956000limx?-3 x2-9x+3=limx? -3x-3x+3x+3=limx? -3x-3x – 3.1 – 3.01 – 3.001 – 2.999 – 2.99 – 2.9

f (X) – 6.1 – 6.01 – 6.001 – 5.999 – 5.99 – 5.9

Example 4 : (Nur Aqilah Binti Adnan)

limx?3×2-9x-3 = limx?3(x-3)(x+3)(x-3) = limx?3x+3Table:

x 2.99 2.999 2.9999 3 3.0001 3.001 3.01

f(x) 5.99 5.999 5.9999 6 6.0001 6.001 6.01

Example 5: (Wan Aisyah Munirah Bt Wan Lokman)

limx?21x-2×2.3 2.2 2.1 2.05 2.01 2.001

f(x)3.333 5 10 20 100 1000

Substitute x=2.3 to get the value of f(x)

limx?21x-2=12.3-2= 3.333

APPLICATION

The concept of the limit, let’s consider a moving car. Suppose we have a car whose position is linear with respect to time (that is, a graph plotting the position with respect to time will show a straight line). We want to find the velocity. This is easy to do from algebra, we just take the slope, and that’s our velocity.

But unfortunately, things in the real world don’t always travel in nice straight lines. Cars speed up, slow down, and generally behave in ways that make it difficult to calculate their velocities.

Now what we really want to do is to find the velocity at a given moment (the instantaneous velocity). The trouble is that in order to find the velocity we need two points, while at any given time, we only have one point. We can find the average speed of the car, given two points in time, but we want to find the speed of the car at one precise moment.

This is the basic trick of differential calculus, we take the average speed at two moments in time, and then make those two moments in time closer and closer together. We then see what the limit of the slope is as these two moments in time are closer and closer, and say that this limit is the slope at a single instant.

CONCLUSION

The concept of the limit of a function is formally introduced. Rules for computing limits are also given, and some situations are described where the limit does not exist.

By the end of our study we should know:

How to evaluate the limit of f(x) as x approaches a number a.

How to evaluate left-hand limits and right-hand limits.

The relationship between the limit of a function on a point, and left-hand and right-hand limits of the function at the point.

How to evaluate the limits of sums, differences, products, and quotients of function.

How to evaluate limits involving absolute value.

REFERENCES

http://en.wikipedia.org/wiki/Limit_(mathematics)#Limit_of_a_function (Concept).

College matriculation mathematics (accounting) QA016,published 2013,MATRICULATION PROGRAMME SYLLABUS,SAP PUBLICATIONS (M) SDN.BHD.(Nur Aqilah ,Norazieatul Wahida,Wan Aisyah).

http://www.math.tamu.edu/AppliedCalc/Workbook/Chapter3.doc. (Nur Aqilah , Norazietul Wahida).

http://www.mathsisfun.com/calculus/limits-evaluating.html (Siti Hajar)

http://en.wikibooks.org/wiki/Calculus/Limits/An_Introduction_to_Limits (Wan Aisyah)

Real Exam Practice Mathematics for Matriculation 1 (Accounting) Based on the 2013/2014syllabus(Siti Nur Fatin, SitiHajar)

http://www.cliffsnotes.com/math/calculus/calculus/limits/evaluating-limits (Siti Nur Fatin)

EXERSICE QUESTIONS?

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