CHAPTER 10 Limits of Trigonometric Functions

1 CHAPTER 10 Limits of Trigonometric FunctionsSome limitsinvolve Trigonometric Functions . This CHAPTER explains howto deal with them. Let s begin with the six Trigonometric of the Six Trigonometric FunctionsWe start with the simple limitlimx!csin(x).Herexis a radian measure becausewe are takingsinof it. And becausethe radian measurexapproacesc,weinterpretcas a radian measure picture on the right illustrates pointxon the unit circle moves to-ward the pointcon the circle. As thishappens,sin(x)approaches the numbersin(c). Thuslimx!csin(x)=sin(c).cxsin(x)sin(c)Fo r example,limx! 4sin(x)=sin 4 =p22. With a slight adaption, the abovepicture also showslimx!ccos(x)=cos(c). And applying limit law 5, we getlimx!ctan(x)=limx!csin(x)cos(x)=limx! csin(x)limx!ccos(x)=sin(c)cos(c)=tan(c), provided thatcos(c)6=0, that is,c6= 2+k , wherekis an integer.

2 In this waythat we get the following !csin(x)=sin(c)for all real numbersclimx!ccos(x)=cos(c)for all real numbersclimx!ctan(x)=tan(c)for all real numbersc6= 2+k limx!csec(x)=sec(c)for all real numbersc6= 2+k limx!ccot(x)=cot(c)for all real numbersc6=k limx!ccsc(x)=csc(c)for all real numbersc6=k 148 Limits of Trigonometric FunctionsExample ! cos(x) the denominator does not approach zero, we can use limit law 5with the rules just derived. Thenlimx! cos(x)x2=limx! cos(x)limx! x2=cos( ) 2= 1 ! /48xtan(x) 2tan(x)4x .Here the denominator approaches zero, so we try to factor and cancel:limx! 48xtan(x) 2 tan(x)4x =limx! 42tan(x) 4x 4x =limx! 42tan(x)=2tan 4 = Squeeze Theorem and Two Important LimitsIt is easy to imagine Limits where factoring and canceling is impossible, orfor which the limit laws do not apply.

3 For example, inlimx!0sin(x)xwe can tfactor anxfrom the top to cancel thexon the bottom (which approaches0).Actually, this particular limit turns out to be significant in calculus. Wenow discuss a theorem that handles Limits such as this one. The idea is tocleverly compare a complicated limit to two simpler (The Squeeze Theorem)Suppose we need to computelimx!cg(x). Suppose also that we can find twofunctionsf(x)andh(x)for whichf(x) g(x) h(x)for values ofxnearc,and for whichlimx!cf(x)=L=limx!ch(x). Thenlimx!cg(x)= (x)y=g(x)y=f(x)cLThe above picture illustrates the squeeze theorem. The graph ofg(x)issqueezed between the graphs off(x)andh(x), both of which approachLasxapproachesc. The squeeze theorem states the obvious fact that in thissituation we can conclude thatg(x) the squeeze theorem to findlimx!

4 Cg(x)requires some have to find two other functionsf(x)andh(x)for whichf(x) g(x) h(x)andbothlimx!cf(x)andlimx!ch(x)are easy to computeandare both equal tothe same numberL. At that point the squeeze theorem sayslimx!cg(x)= Squeeze Theorem and Two Important Limits149We will next use the squeeze theorem to findlimx!0sin(x)x, which will beneeded in CHAPTER 21. But first let s think about what we d expect it to unit circle on the right shows a radian mea-surex, close to0. the vertical side of the tri-angle is the corresponding valuesin(x). Bothsin(x)andxare small, but the curved arcxis so small that it looks almost like a verticalline. The smallerx, the more vertical it looks,and in fact it becomes almost indistinguishablefrom the vertical sidesin(x). For very smallxthe ratiosin(x)xappears to be quite close might guesslimx!

5 0sin(x)x= (x)In fact, this turns out to be exactly the case. Proving it with the squeezetheorem requires a formula from geometry. Recall that asector of a circleis a pie slice of the circle, as illustrated below, :The area of a sector of a circle ofanglexand radiusrisA= is why the formula works: The area of acircle of radiusris r2. The sector takes uponly a fraction of this circle, that fraction beingxradians out of2 radians around the entireunit circle, orx2 . ThusA= r2 x2 = are ready to carry out our plan of provingthatlimx!0sin(x)x=1via the squeeze theorem. Wewill concoct two functionsf(x)andh(x)withf(x) sin(x)x h(x)andlimx!0f(x)=1=limx!0h(x).The functionsfandhwill come from the dia-gram on the right showing a sectorOCPon theunit circle, and another sectorOABof radiuscos(x)inside it. From this we get the following:x|{z}cos(x)sin(x)8<:OCPBA150 Limits of Trigonometric Functions Area ofsectorOAB!

6 Area oftriangleOCP! Area ofsectorOCP!.Using the area formula for a sector (from the previous page) and the areaformula for a triangle (from heart), this becomes12 cos2(x) x 12 1 sin(x) 12 12 , this only works ifxispositive. If it were negative, then the above areas would be negative too. We correct this by taking the absolute valueof the potentially negative termsxandsin(x).12 cos2(x) |x| 12 1 |sin(x)| 12 12 |x|.Now multiply all parts of this inequality by the positive number2|x|to getcos2(x) |sin(x)||x| this point the absolute values are unnecessary because ifxis close tozero (as it is whenx!0), thenxandsin(x)are either both positive or bothnegative, sosin(x)xis already positive. Updating the above, we getcos2(x) sin(x)x we ve squeezedy=sin(x)xbetween the functionsy=cos2(x)andy= (x)y=sin(x)xy=cos2(x)0 Becauselimx!

7 0cos2(x)=cos2(0)=1=limx!01, the squeeze theorem guaranteeslimx!0sin(x)x=1.( )From this day forward, remember the fundamental factlimx!0sin(x)x= Squeeze Theorem and Two Important Limits151 Below is a more complete picture of this situation, showingy=sin(x)xwithy=cos2(x)andy=1. Notice that it s not the case thatcos2(x) sin(x)x 1foreveryvalue ofx. But thisdoeshold whenxis near zero, and that is all weneeded to apply the squeeze (x)x 2 3 4 5 2 3 4 5 Students often assert incorrectly thatsin(x)x=1. But that plainly above graph shows thatsin(x)xneverequals1. In fact,sin(x)x<1for anyxexcept0, and it is undefined whenx=0. What we have determined is thatit grows ever closer to1asxapproaches zero, that is,limx!0sin(x)x= we use this fact to compute another significant !0cos(x) course we can t just plug inx=0because that would give can we factor anything from the top to cancel with thexon the let s entertain a little wishful thinking.

8 If we could only change the topfromcos(x) 1tocos2(x) 1, then the identitysin2(x)+cos2(x)=1would turnthe top intocos2(x) 1= sin(x), and we d get our familiar formsin(x) accomplish just this by multiplying by the conjugate ofcos(x) !0cos(x) 1x=limx!0cos(x) 1x cos(x)+1cos(x)+1 multiply by1=cos(x)+1cos(x)+1=limx!0cos2(x) 1x(cos(x)+1) FOIL top=limx!0 sin2(x)x(cos(x)+1) usecos2(x) 1= sin2(x)=limx!1sin(x)x sin(x)cos(x)+1 regroup=limx!0sin(x)x limx!0 sin(x)cos(x)+1 apply limit laws=1 sin(0)cos(0)+1 uselimx!0sin(x)x=1, limit laws=1 01+1=0 final answer!Thereforelimx!0cos(x) 1x= of Trigonometric FunctionsHere is a summary of what we developed over the previous three Limits will be useful later, and should be (Two Important Limits )limx!0sin(x)x=1limx!0cos(x) 1x=0 These (especially the first) are useful for finding various other !

9 0tan(x) insertingx=0results intan(0)0=00, so we try a di erent approach:limx!0tan(x)x=limx!0sin(x)cos(x )x=limx!0sin(x)cos(x)x1=limx!0sin(x)cos( x) 1x=limx!0sin(x)x 1cos(x)=1 1cos(0)= a limit will not have the exact form of one in Theorem can be made to match with a little algebra. For instance, let s work outlimx!0sin(2x)x. Here the2xin the sin is not the same as thexon the bottom,so this does not exactly match the familiarlimx!0sin(x)x=1. To see how to fixthis it s helpful to emphasize the structure of this limit by replacing thexwith a boxthat could represent any expression:lim!0sin()= meanssin()approaches 1 as the gray box approaches 0. Look againat the limit we re trying to evaluate:limx!0sin(2x)x. Thexon the bottom doesnot match the2xin the box. But we can make it match by multiplying thefraction by1=22and factoring out the 2 on the top:limx!

10 0sin(2x)x=limx!022 sin(2x)x=2 limx!0sin(2x)2x=2 1= !0sin(2x)2x=1becausex!0makes2x!0, hencesin(2x)2x!1. Inconclusionlimx!0sin(2x)x= Squeeze Theorem and Two Important Limits153 Note that the following approach (which gives the correct answer) iswrong becausesin(2x)6=2sin(x).limx!0sin(2x)x=l imx!02sin(x)x=2limx!0sin(x)x=2 1= !3sin(h 3)h2+2h dividing the limit of the top by the limit of the bottom results insin(3 3)32+2 3 15=00, so we have to try something else. factoring gives a match:limh!3sin(h 3)h2+2h 15=limh!3sin(h 3)(h+5)(h 3)=limh!31h+5sin(h 3)h Theorem !3sin(h 3)h 3=1, becauseh 3approaches 0 ash! the above calculation,limh!31h+5sin(h 3)h 3=limh!31h+5 limh!3sin(h 3)h 3=18 1= !3sin(h 3)h2+2h 15= !0sin( x)sin(3x).Blindly trying a limit law yieldslimx!0sin( x)sin(3x)=limx!0sin( x)limx!0sin(3x)=sin(0)sin(0)=00, sowe need to follow a di erent path.