Lecture 4 : Calculating Limits using Limit Laws

1 Lecture 4 : Calculating Limits using Limit LawsClick on this symbolto view an interactive demonstration in Wolfram the definition of the Limit , limx af(x), we can derive many general laws of Limits , that help us tocalculate Limits quickly and easily. The following rules apply to any functionsf(x) andg(x) and alsoapply to left and right sided Limits :Suppose thatcis a constant and the limitslimx af(x)andlimx ag(x)exist (meaning they are finite numbers).Then1. limx a[f(x) +g(x)] = limx af(x) + limx ag(x) ;(the Limit of a sum is the sum of the Limits ).2. limx a[f(x) g(x)] = limx af(x) limx ag(x) ;(the Limit of a difference is the difference of the Limits ).3. limx a[cf(x)] =climx af(x);(the Limit of a constant times a function is the constant times the Limit of the function).

2 4. limx a[f(x)g(x)] = limx af(x) limx ag(x);(The Limit of a product is the product of the Limits ).5. limx af(x)g(x)=limx af(x)limx ag(x)if limx ag(x)6= 0;(the Limit of a quotient is the quotient of the Limits provided that the Limit of the denominator isnot 0)ExampleIf I am given thatlimx 2f(x) = 2,limx 2g(x) = 5,limx 2h(x) = the Limits that exist (are a finite number):(a) limx 22f(x) +h(x)g(x)=limx 2(2f(x) +h(x))limx 2g(x)since limx 2g(x)6= 0=2 limx 2f(x) + limx 2h(x)limx 2g(x)=2(2) + 05=45(b) limx 2f(x)h(x)(c) limx 2f(x)h(x)g(x)Note 1If limx ag(x) = 0 and limx af(x) =b, wherebis a finite number withb6= 0, Then:the values of the quotientf(x)g(x)can be made arbitrarily large in absolute value asx aand thus1the Limit does not the values off(x)g(x)are positive asx ain the above situation, then limx af(x)g(x)= ,If the values off(x)g(x)are negative asx ain the above situation, then limx af(x)g(x)= ,If on the other hand, if limx ag(x) = 0 = limx af(x), we cannot make any conclusions aboutthe limx cosxx.

3 Asxapproaches from the left, cosxapproaches a finite number from the left,x approaches asxapproaches from the left, the quotientcosxx approaches in absolute values of both cosxandx are negative asxapproaches from the left, thereforelimx cosxx = .More powerful laws of Limits can be derived using the above laws 1-5 and our knowledge of somebasic functions. The following can be proven reasonably easily ( we are still assuming thatcis aconstant and limx af(x) exists );6. limx a[f(x)]n=[limx af(x)]n, wherenis a positive integer (we see this using rule 4 repeatedly).7. limx ac=c, where c is a constant ( easy to prove from definition of Limit and easy to see fromthe graph,y=c).

4 8. limx ax=a, (follows easily from the definition of Limit )9. limx axn=anwherenis a positive integer (this follows from rules 6 and 8).10. limx an x=n a, where n is a positive integer anda >0 if n is even. (proof needs a little extrawork and the binomial theorem)11. limx an f(x) =n limx af(x) assuming that the limx af(x)>0 ifnis even. (We will look atthis in more detail when we get to continuity)ExampleEvaluate the following Limits and justify each step:(a)limx 3x3+2x2 x+1x 1(b)limx 13 x+ 12(c)Determine the infinite Limit (see note 1 above, say if the Limit is , or )limx 2 x+1(x 2).Polynomial and Rational FunctionsPlease review the relevant parts ofLectures 3, 4 and 7from the Algebra/Precalculus reviewpage.

5 This demonstration will help you visualize some rational functions:Direct Substitution (Evaluation) PropertyIffis a polynomial or a rational function andais in the domain off, then limx af(x) =f(a). This follows easily from the rules shown above.(Note that this is the case in part (a) of the example above)iff(x) =P(x)Q(x)is a rational function whereP(x) andQ(x) are polynomials withQ(a) = 0, then:IfP(a)6= 0, we see from note 1 above that limx aP(x)Q(x)= or and is not equal to .IfP(a) = 0we can cancel a factor of the polynomialP(x) with a factor of the polynomialQ(x)and the resulting rational function may have a finite Limit or an infinite Limit or no Limit atx= Limit of the new quotient asx ais equal to limx aP(x)Q(x)by the following observation whichwe made in the last Lecture :Note 2: Ifh(x) =g(x) whenx6=a, then limx ah(x) = limx ag(x) provided the Limits if the following Limits are finite, equal to or and are not equalto :(a) limx 3x2 9x 3.

6 (b)limx 1 x2 x 6x 1.(c)Which of the following is true:1. limx 1x2 x 6x 1= + ,2. limx 1x2 x 6x 1= ,3. limx 1x2 x 6x and is not ,3 ExampleEvaluate the Limit (finish the calculation)limh 0(3 +h)2 (3) 0(3+h)2 (3)2h= limh 09+6h+h2 9h=ExampleEvaluate the following Limit :limx 0 x2+ 25 also our observation from the last day which can be proven rigorously from the definition(this is good to keep in mind when dealing with piecewise defined functions):Theormlimx af(x) =Lif and only iflimx a f(x) =L= limx a+f(x).ExampleEvaluate the Limit if it exists:limx 23x+ 6|x+ 2|The following theorems help us calculate some important Limits by comparing the behavior of afunction with that of other functions for which we can calculate Limits :4 TheoremIff(x) g(x) whenxis near a(except possible ata) and the Limits off(x) andg(x)both exist asxapproachesa, thenlimx af(x) limx ag(x).

7 The Sandwich (squeeze) TheoremIff(x) g(x) h(x) whenxis neara(exceptpossibly ata) andlimx af(x) = limx ah(x) =Lthenlimx ag(x) = last day, we saw that limx 0sin(1/x) does not exist because of how the function oscil-lates nearx= 0. However we can see from the graph below and the above theorem thatlimx 0x2sin(1/x) = 0, since the graph of the function is sandwiched betweeny= x2andy=x2:O O the Limit limx have 1 sin(1/x) 1 for allx,multiplying across byx2(which is positive), we get x2 x2sin(1/x) x2for allx, using the Sandwich theorem, we get0 = limx 0 x2 limx ox2sin(1/x) limx 0x2= 0 Hence we can conclude thatlimx 0x2sin(1/x) = if the following Limit exists and if so find its values:limx ox100cos2( /x)5 Extra Examples, attempt the problems before looking at the solutionsDecide if the following Limits exist and if a Limit exists, find its value.

8 (1) limx 1x4+ 2x3+x2+ 3(2) limx 2x2 3x+2(x 2)2.(3) limx 0(1x 1|x|).(4) limx 0|x|x2+x+10.(5) limh 0 4+h 2h.(6) If 2x g(x) x2 x+ 2 for allx, evaluate limx 1g(x).(7) Determine if the following Limit is finite, or and is not .limx 1 (x 3)(x+ 2)(x 1)(x 2).6 Extra Examples, attempt the problems before looking at the solutionsDecide if the following Limits exist and if a Limit exists, find its value.(1) limx 1x4+ 2x3+x2+ 3 Since this is a polynomial function, we can calculate the Limit by direct substitution:limx 1x4+ 2x3+x2+ 3 = 14+ 2(1)3+ 12+ 3 = 7.(2) limx 2x2 3x+2(x 2) is a rational function, where both numerator and denominator approach 0 as x approaches2.

9 We factor the numerator to getlimx 2x2 3x+ 2(x 2)2= limx 2(x 1)(x 2)(x 2)2 After cancellation, we getlimx 2(x 1)(x 2)(x 2)2= limx 2(x 1)(x 2).Now this is a rational function where the numerator approaches 1 asx 2 and the denominatorapproaches 0 asx 2. Thereforelimx 2(x 1)(x 2)does not can analyze this Limit a little further, by checking out the left and right hand Limits at 2. Asxapproaches 2 from the left, the values of (x 1) are positive (approaching a constant 1) andthe values of (x 2) are negative ( approaching 0). Therefore the values of(x 1)(x 2)are negative andbecome very large in absolute value. Thereforelimx 2 (x 1)(x 2)= .Similarly, you can show thatlimx 2 (x 1)(x 2)= + ,and therefore the graph ofy=(x 1)(x 2)has a vertical asymptote atx= 2.

10 (check it out on your calculator)(3) limx 0(1x 1|x|).Letf(x) =1x 1|x|. We write this function as a piecewise defined function:f(x) = 1x 1x= 0x >01x+1x=2xx 0(1x 1|x|) exists only if the left and right hand Limits exist and are 0+(1x 1|x|) = limx 0+0 = 0 and limx 0 (1x 1|x|) = limx 0 2x= .Since the Limits do not match, we havelimx 0(1x 1|x|) (4) limx 0|x|x2+x+ limx 0x2+x+ 10 = 106= 0, we havelimx 0|x|x2+x+ 10=limx 0|x|limx 0(x2+x+ 10)=limx 0|x| |x|= x x >0 x x limx 0+|x|= 0 = limx 0 |x|. Hence limx 0|x|= 0 andlimx 0|x|x2+x+ 10=limx 0|x|10=010= 0.(5) limh 0 4+h limh 0 4 +h 2 = 0 = limh 0h, we cannot determine whether this Limit exists or notfrom the Limit laws without some transformation.