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CHAPTER 10Limits of Trigonometric FunctionsSome limits involve trigonometric functions. This Chapter explains howto deal with them. Let’s begin with the six trigonometric functions.10.1 Limits of the Six Trigonometric FunctionsWe start with the simple limit lim sin(x).x! cHere x is a radian measure becausewe are taking sin of it. And becausethe radian measure x approaces c, weinterpret c as a radian measure too.The picture on the right illustrates this.The point x on the unit circle moves toward the point c on the circle. As thishappens, sin(x) approaches the numbersin(c). Thus lim sin(x) sin(c).x! ccxsin(x) sin(c)p2For example, limº sin(x) sin . With a slight adaption, the above42x! 2picture also shows lim cos(x) cos(c). And applying limit law 5, we get º x! clim sin(x)sin(x)sin(c)x! c tan(c),x! c cos(x)lim cos(x)cos(c)lim tan(x) limx! cx! cprovided that cos(c) 6 0, that is, c 6 º2 kº, where k is an integer. In this waythat we get the following formulas.lim sin(x) sin(c)x! clim cos(x) cos(c)x! clim tan(x) tan(c)x! clim sec(x) sec(c)x! clim cot(x) cot(c)x! clim csc(x) csc(c)x! cfor all real numbers cfor all real numbers cfor all real numbers c 6 º2 kºfor all real numbers c 6 º2 kºfor all real numbers c 6 kºfor all real numbers c 6 kº

148Limits of Trigonometric FunctionsExample 10.1Find limx!ºcos(x).x2Because the denominator does not approach zero, we can use limit law 5with the rules just derived. Then limx!ºExample 10.2 Find limx!º/4cos(x) cos(º) 1cos(x) xlim!º 2.x2º2ºlim x28x tan(x) 2 tan(x).4x ºx !ºHere the denominator approaches zero, so we try to factor and cancel: º 2 tan(x) 4x º8x tan(x) 2º tan(x)limº limº limº 2 tan(x) 2 tan 2.x! 44x º4x ºx! 44x! 410.2 The Squeeze Theorem and Two Important LimitsIt is easy to imagine limits where factoring and canceling is impossible, orsin( x)we can’tx!0xfor which the limit laws do not apply. For example, in limfactor an x from the top to cancel the x on the bottom (which approaches 0).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 limits.Theorem 10.1 (The Squeeze Theorem)Suppose we need to compute lim g(x). Suppose also that we can find twox! cfunctions f (x) and h(x) for which f (x) g(x) h(x) for values of x near c,and for which lim f (x) L lim h(x). Then lim g(x) L.x! cx! cx! cyy h( x)y g ( x)Ly f ( x)cxThe above picture illustrates the squeeze theorem. The graph of g(x) issqueezed between the graphs of f (x) and h(x), both of which approach L asx approaches c. The squeeze theorem states the obvious fact that in thissituation we can conclude that g(x) approaches L too.Applying the squeeze theorem to find lim g(x) requires some ingenuity.x! cWe have to find two other functions f (x) and h(x) for which f (x) g(x) h(x)and both lim f (x) and lim h(x) are easy to compute and are both equal tox! cx! cthe same number L. At that point the squeeze theorem says lim g(x) L.x! c

149The Squeeze Theorem and Two Important LimitsWe will next use the squeeze theorem to find limx!0sin( x), which will bexneeded in Chapter 21. But first let’s think about what we’d expect it to be.The unit circle on the right shows a radian measure x, close to 0. the vertical side of the triangle is the corresponding value sin(x). Bothsin(x) and x are small, but the curved arc xis so small that it looks almost like a verticalline. The smaller x, the more “vertical” it looks,and in fact it becomes almost indistinguishablefrom the vertical side sin(x). For very small xsin( x) xsin( x)appears to be quite close to 1.xsin( x)We might guess lim 1.x !0xthe ratioIn fact, this turns out to be exactly the case. Proving it with the squeezetheorem requires a formula from geometry. Recall that a sector of a circleis a “pie slice” of the circle, as illustrated below, shaded.Formula: The area of a sector of a circle ofangle x and radius r isA 1 2r x.2rHere is why the formula works: The area of acircle of radius r is º r 2 . The sector takes uponly a fraction of this circle, that fraction beingx radians out of 2º radians around the entirexx1unit circle, or. Thus A º r 2 · r 2 x.2º2ºxA12We are ready to carry out our plan of provingsin( x) 1 via the squeeze theorem. Wex !0xthat limwill concoct two functions f (x) and h(x) withsin(x)f (x) h(x) and lim f (x) 1 lim h(x).x!0x!0xThe functions f and h will come from the diagram on the right showing a sector OCP on theunit circle, and another sector O AB of radiuscos(x) inside it. From this we get the following:sin( x)8 :OPB {z } Acos( x)xC

150Limits of Trigonometric Functions Area ofsector O AB! Area oftriangle OCP! Area ofsector OCP!.Using the area formula for a sector (from the previous page) and the areaformula for a triangle (from heart), this becomes111 2· cos2 (x) · x · 1 · sin(x) · 1 · x.222Actually, this only works if x is positive. If it were negative, then the above“areas” would be negative too. We correct this by taking the absolute valueof the potentially negative terms x and sin(x).111 2· cos2 (x) · x · 1 · sin(x) · 1 · x .222Now multiply all parts of this inequality by the positive number2 x to get sin(x) 1. x cos2 (x) At this point the absolute values are unnecessary because if x is close tozero (as it is when x ! 0), then x and sin(x) are either both positive or bothnegative, so sin(x x) is already positive. Updating the above, we getsin(x) 1.xcos2 (x) Now we’ve squeezed y sin(x x) between the functions y cos2 (x) and y 1.yy 1y y cos2 ( x)0sin( x)xy cos2 ( x)Because lim cos2 (x) cos2 (0) 1 lim 1, the squeeze theorem guaranteesx !0x!0sin(x) 1.x!0x(10.1)limsin(x) 1.x!0xFrom this day forward, remember the fundamental fact lim

151The Squeeze Theorem and Two Important LimitsBelow is a more complete picture of this situation, showing y sin(x x) withy cos2 (x) and y 1. Notice that it’s not the case that cos2 (x) sin(x x) 1 forevery value of x. But this does hold when x is near zero, and that is all weneeded to apply the squeeze theorem.1x 5º 4º 3º 2º ºy sin( x)xº2º3º4º5ºxStudents often assert incorrectly that sin(x x) 1. But that plainly wrong.The above graph shows that sin(x x) never equals 1. In fact, sin(x x) 1 for any xexcept 0, and it is undefined when x 0. What we have determined is thatit grows ever closer to 1 as x approaches zero, that is, limx !0sin( x) 1.xNow we use this fact to compute another significant limit.cos(x) 1.x!0xOf course we can’t just plug in x 0 because that would give the 00 nonsense.Nor can we factor anything from the top to cancel with the x on the bottom.Example 10.3 Find limBut let’s entertain a little wishful thinking. If we could only change the topfrom cos(x) 1 to cos2 (x) 1, then the identity sin2 (x) cos2 (x) 1 would turnthe top into cos2 (x) 1 sin(x), and we’d get our familiar form sin(x x) . Wecan accomplish just this by multiplying by the conjugate of cos(x) 1.limx !0cos(x) 1x Therefore limx !0cos(x) 1 cos(x) 1·x!0xcos(x) 1cos2 (x) 1limx!0 x(cos(x) 1) sin2 (x)limx!0 x(cos(x) 1)sin(x) sin(x)lim·x!1xcos(x) 1sin(x) sin(x)lim· limx!0x!0 cos(x) 1xsin(0)1·cos(0) 101· 01 1limcos( x) 1 0.x multiply by 1 cos( x) 1cos( x) 1 FOIL top use cos2 ( x) 1 sin2 ( x) regroup apply limit lawssin( x) 1, limit lawsx !0x use lim final answer!

152Limits of Trigonometric FunctionsHere is a summary of what we developed over the previous three pages.These limits will be useful later, and should be remembered.Theorem 10.2limx!0(Two Important Limits)sin(x) 1xlimx !0cos(x) 1 0xThese (especially the first) are useful for finding various other limits.Example 10.4 Find limx !0tan(x).xJust inserting x 0 results intan(x)lim limx !0x!0xsin( x)cos( x)x 00 , so we try a different approach:tan(0)0sin( x)cos( x)lim xx!01 limx !0sin(x) 1·cos(x) xsin(x)1·x!0xcos(x)1 1· 1.cos(0) limSometimes a limit will not have the exact form of one in Theorem 10.2but can be made to match with a little algebra. For instance, let’s work outsin(2x). Here the 2x in the sin is not the same as the x on the bottom,x!0xsin(x)so this does not exactly match the familiar lim 1. To see how to fixx !0xthis it’s helpful to emphasize the structure of this limit by replacing the xlimwith a boxthat could represent any expression:limsin( )!0This meanssin( ) 1.approaches 1 as the gray box approaches 0. Look againsin( 2x ). The x on the bottom doesx!0xat the limit we’re trying to evaluate: limnot match the 2x in the box. But we can make it match by multiplying thefraction by 1 22 and factoring out the 2 on the top:sin( 2x )2 sin( 2x )sin( 2x ) lim · 2 lim 2 · 1 2.x!0x!0 2x !0xx2xlimHere limx!0sin( 2x )2xconclusion limx!0 1 because x ! 0 makes 2x ! 0, hencesin(2x) 2.xsin( 2x )2x! 1. In

The Squeeze Theorem and Two Important Limits153Note that the following approach (which gives the correct answer) iswrong because sin(2x) 6 2 sin(x).sin(2x)2 sin(x)sin(x) lim 2 lim 2 · 1 2.x!0x!0x!0xxxlimExample 10.5 Find limh !3sin(h 3).h2 2h 15Just dividing the limit of the top by the limit of the bottom results insin(3 3)32 2·3 150 , so we have to try something else. Factoring gives a match:0sin(h 3)sin(h 3)1 sin( h 3 ) lim lim.2h!3 h 2h 15h!3 (h 5)(h 3)h !3 h 5h 3limBy Theorem 10.2 limh!3sin( h 3 )h 3 1, because h 3 approaches 0 as h ! 3.Continuing the above calculation,1 sin( h 3 )1sin( h 3 )11 lim· lim ·1 .h !3 h 5h !3 h 5 h !388h 3h 3limTherefore limh !3sin(h 3)1 .2h 2h 15 8Example 10.6Find limx !0sin(º x).sin(3x)Blindly trying a limit law yields limx!0lim sin(º x)sin(º x)sin(0)0x!0 , sosin(3x)lim sin(3x)sin(0)0x!0we need to follow a different path. This one gives a match to Theorem 10.2:sin(º x)sin(º x)limsin(º x)º sin(º x) 3xººº 1x!0ºxºxlim lim lim · .sin(3x)x!0 sin(3x)x !0 3º x sin(3x)3 x!0 sin(3x)33 1limx!03x3xTherefore limx !0sin(º x) º .sin(3x) 3If you can work the exercises you are ready to move on.

154Limits of Trigonometric FunctionsExercises for Chapter 10Find the limits.1. limº cot(x)x!º2. lim cos(x)13. limx!5º4. lim cosx!714. lim ºx µ65º5. lim tanx!3xcos2 (x)x!º/2 sin3. limº tan(h)h! 4x cos(x) xcos(x) 112. limx! 2x!º/215. lim 2(x) 1º sin x2xº sin hh !02h1 cos(µ )6. limº (cos(x) 2 cot(x)) 16. µlim!0 sin(2µ )x!47. limº (7 2 cos(x))23x! 3cos2 (x) cos(x)8. limx!ºcos(x) 1p9. lim 3 tan(x) 1x!º/410. lim cos(x) sin(x)x! 74ºcos2 (x) cos(x)11. limx!0cos(x) 1sin(2x 4)17. limx!2 5x 10tan xx!0 3x18. limsin(2x 2)19. limx!1x 12x20. limx!0 sin(3x)sin(x)21. limx!0 sin(2x)tan(º x)x !02x22. limsin(x2 1)x !1x 123. lim24. lim1tan(3µ )µ25. lim1µ cot(4µ )26. limsec(µ )µ csc(2µ )µ !0µ !0µ !01µ27. limcos(µ )µ !028. limx !029. limsin(5µ )sin(7x)5xx!2ºsin(1 cos x)1 cos xpsin( 9x)30. limpx!0x31. In Theorem 10.2 we interpret x as a radian measure. Show that if x weredegrees (not radians) then limx !0sin(x)º . This is yet another reason wex180prefer radians over degrees. By measuring angles with radians, limhas the simplest possible value, namely 1.x !0sin( x)x

155Exercise Solutions for Chapter 1010.3 Exercise Solutions for Chapter 101. limº cot(x) cot(º/2) 0x! 25. lim tanx!3µ µ3. limº tan(h) tan(º/4) 1 p5º5º tan 3x3µ237. lim (7 2 cos(x)) lim (7 2 cos(x))x!º/39. limx!º/4p3 tan(x) 1 x!º/3qh! 4 23 lim (3 tan(x) 1) x!º/4pp22337 2 cos(º/3) 8 4p3 tan(º/4) 1 p3 1 2cos2 (x) cos(x)cos(x) (cos(x) 1) lim lim cos(x) 1x!0x !0x!0cos(x) 1cos(x) 111. limcos2 (x)13. limx!º/2 sin2 (x) 1cos2 (x) 1x!º/2 cos2 (x) lim(because sin2 (x) 1 cos2 (x))15. limº sin h17. limsin(2x 4)sin(2x 4)2 sin(2x 4) 2sin(2x 4) 2 lim lim lim x!2 5(x 2)x!2 2 · 5(x 2)5x 105 x!2 2x 452hh !0x !219. limx !1 sin h ºº ·1 2 h !0 h22ºlimsin(2x 2)2 sin(2x 2)sin(2x 2) lim 2 lim 2·1 2x!1x!1x 12(x 1)2x 2sin(x)sin(x) 2x1sin(x) 2x121. lim lim lim limx!0 sin(2x)x!0 2x sin(2x)x!02x sin(2x) 2 x!0sin( x)xsin(2 x)2x 1 1 1· 2 1 2 sin (x 1)(x 1)(x 1) sin (x 1)(x 1)sin(x2 1)23. lim lim limx !1x!1x!1x 1x 1(x 1)(x 1) lim (x 1)x!125. limµ !027. limµ !029. limx!2ºsin(x2 1)sin(x2 1) lim(x 1)lim (1 1) · 1 2x !1x !1x2 1x2 11tan(4µ )sin(4µ )4sin(4µ ) 4 lim lim lim ·1 4µ !0 µ cos(4µ )µ !0 cos(4µ )µ cot(4µ ) µ!0µ4µ11µsin(5µ )cos(µ ) limµ !05 sin(5µ ) 5 ·1 5cos(µ ) 5µ1sin(1 cos x) 1 (because 1 cos x ! 0 as x ! 2º)1 cos x

The Squeeze Theorem and Two Important Limits 151 Below is a more complete picture of this situation, showing y sin(x) x with y cos2(x) and y 1.Notice that it’s not the case that cos2(x) sin(x) x 1 for every value of x.But this does hold when is near zero