The Squeeze Theorem - UCLA Mathematics

1 Math 31A Discussion SessionWeek 2 NotesJanuary 12 and 14, 2016 This week we ll discuss a powerful tool for computing limits, called the Squeeze theo-rem. Following this, we may also mention limits at infinity, whose computation sometimesrequires different methods. Finally, we will give a geometric motivation for the derivative,and investigate some of its Squeeze TheoremAs useful as the limit laws are, there are many limits which simply will not fall to thesesimple rules. One helpful tool in tackling some of the more complicated limits is theSqueezeTheorem: Theorem ,g, andhare functions so thatf(x) g(x) h(x)neara, with the exception that this inequality might not hold whenx=a. Thenlimx af(x) limx ag(x) limx ah(x),if these three limits exist. In particular, if limx af(x) =L= limx ah(x), thenlimx ag(x) = Evaluate the limit limx 0(x cos(1/x)), if it exists.(Solution) We know that 1 cos(1/x) 1 for allx6= 0.

2 Then x x cos(1/x) x,solimx 0( x) limx 0(x cos(1/x)) limx limx 0( x) = 0 = limx 0x, we see thatlimx 0(x cos(1/x)) = Use the Squeeze Theorem to evaluate lim 0(sin ), if it exists.(Solution) The following figure will prove to be useful in evaluating this limit:1In the above figure, the blue curve is the portion of the unit circle which lies in the firstquadrant, and the orange ray makes an angle of with the origin, where 0< < green line is the linex= 1, so the intersection of the orange and green lines is thepoint (1,tan( )). Consider the right triangle made by the orange line, the green line,and thex-axis; this triangle has height tan and width 1, so its area isA1=tan consider the sector of the unit circle that lies between thex-axis and the orangeray. This sector has angle , so its area isA2= 2 = , consider the triangle formed by the orange line, the purple line, and triangle has height sin and width 1, so its area isA3=sin since this triangle lies entirely inside the sector, and the sector lies entirely insidethe first triangle, we haveA1 A2 A3 tan 2 2 sin can multiply through this inequality by 2; taking the reciprocal of each part willthen reverse the inequality:cos sin 1 1sin.

3 Finally, multiply through by sin to obtaincos sin 1.(1)2(Since 0< < 2, sin >0, so the inequality will not reverse.) We have only seenthat inequality (1) holds when 0< < 2, but a similar argument will show that italso holds when 2< <0, so, since lim 0cos = 1 and lim 01 = 1, the SqueezeTheorem allows us to conclude thatlim 0sin = Use the previous example to evaluatelimx 0sin(4x)sin(6x),if this limit exists.(Solution) Forx6= 0 we can rewrite this quotient assin(4x)sin(6x)=x sin(4x)x sin(6x)=(46)(6xsin(6x))(sin(4x)4x).Thenl imx 0sin(4x)sin(6x)=46(limx 06xsin(6x))(limx 0sin(4x)4x)=46(limx 0sin(6x)6x) 1(limx 0sin(4x)4x)=46 1 1 = at InfinityWe ll carry out two illustrative examples of limits at Find limx 8x5+ 3x2 44 9x5, if it exists.(Solution) Neither limx (8x5+ 3x2 4) nor limx (4 9x5) exists, so we cannotvery well consider a ratio of these limits. What we can do, however, is rewrite thisquotient so that the numerator and denominator limits exist.

4 Forx6= 0, we have8x5+ 3x2 44 9x5=8x5+ 3x2 44 9x5 x 5x 5=8 + 3x 3 4x 54x 5 9,3solimx 8x5+ 3x2 44 9x5= limx 8 + 3x 3 4x 54x 5 9=limx (8 + 3x 3 4x 5)limx (4x 5 9)=8 9= technique of writing the denominator as a constant term plus terms with negativeexponents is a good general strategy for determining the end behavior of Considerf(x) =sin(2x+ 7) cos(x2) + cos2(4 x3)x. Find limx f(x), if this limitexists.(Solution) This limit may look daunting, but we need only recall that the sine andcosine functions are bounded. Since sine and cosine take values between 1 and 1, thevalues of the product sin(2x+ 7) cos(x2) will be between 1 and 1. That is, 1 sin(2x+ 7) 1 and 1 cos(x2) 1,so 1 sin(2x+ 7) cos(x2) , since 1 cos(4 x3) 1, 0 cos2(4 x3) 1. Adding these inequalitiestogether, 1 sin(2x+ 7) cos(x2) + cos2(4 x3) these inequalities overx, we have 1x sin(2x+ 7) cos(x2) + cos2(4 x3)x the terms on each end will tend to zero asxtends to , our limit is DerivativeGiven a continuous functionf, suppose we want to find the equation of the line which liestangent to the graph ofy=f(x) at the point (a,f(a)).

5 At first glance this seems like adifficult problem to approach naively (that is, without derivatives). But we can tackle itwith relative ease by consideringsecant a continuous functionf, thesecant linepassing through (a,f(a)) and(b,f(b)) (wherea6=b) to the graph ofy=f(x) is defined byy=f(b) f(a)b a(x a) +f(a).(2)Intuitively, we can see that the secant line passing through (a,f(a)) and (b,f(b)) shouldbegin to look more and more like the tangent line at (a,f(a)) asbgets closer toa. So weshould be able to obtain the equation of the tangent line by taking the limit of equation (2). the equation of the tangent line to the graph ofy=x2+ 4 at (3,13).(Solution) For someb6= 3, the equation of the secant line passing through (3,13) and(b,f(b)) is given byy=f(b) f(3)b 3(x 3) + 13 =(b2+ 4) 13b 3(x 3) + 13.(3)Notice that the only part of this equation which depends onbis the slope,b2 9b 3.

6 So weshould be able to find the tangent line by finding the limit of this slope asbapproaches 3,and replacing the slope in (3) with this limit. We havelimb 3b2 9b 3= limb 3(b 3)(b+ 3)b 3= limb 3(b+ 3) = 6,so the equation of the tangent line is given byy= 6(x 3) + 13 = 6x , we re just interested in theslopeof the tangent line to a curve. To determinethis, first notice that the slope of the line determined in (2) is given byf(b) f(a)b the slope of the tangent line should be the slope we obtain by lettingbget close toa:limb af(b) f(a)b way to consider this is as follows. Suppose we want to find the slope of the tangentline toy=f(x) at the point (x,f(x)). If we change ourx-value by a small amount say,byh then the coordinates of the point on the graph will be (x+h,f(x+h)). So the slopeof the secant line between this two points isf(x+h) f(x)(x+h) x=f(x+h) f(x) before, we want to see what happens to this quantity as our second point approaches ourfirst.

7 That is, we want to see what happens when our displacementhgets closer to 0, so weconsiderlimh 0f(x+h) f(x) this limit exists, we call the resulting value thederivativeoffatx, and denote thisvariously asddx(f(x)), f (x), f(x). the derivative off(x) =x3at an arbitrary pointx.(Solution) According to our above definition, the derivative will be given byf (x) = limh 0f(x+h) f(x)h= limh 0(x+h)3 x3h= limh 0x3+ 3hx2+ 3h2x+h3 x3h= limh 03hx2+ 3h2x+h3h= limh 0(3x2+ 3hx+h2) = 3x2,whenever this limit exists. But this limit exists for allx-values, sofis everywhere differen-tiable, andf (x) = 3x2for thing we notice immediately is that constant functions have derivative 0. To seethis, notice that ifg(x) =c, thenlimh 0g(x+h) g(x)h= limh 0c ch= limh 00h= fits with our understanding of the derivative as an instantaneous slope, since the graphof a constant function is a horizontal line. Because it is defined as a limit, the derivative alsohas the following pleasant arithmetic properties:Proposition (x) andg (x) exist andc,dare real numbers.

8 (cf(x) +dg(x)) =cf (x) +dg (x); (cf(x) dg(x)) =cf (x) dg (x).The derivatives of products and quotients aren t quite as straightforward, but we ll discussthem soon. Finally, we introduce what we ll call thepower rulefor computing (x) =xnfor a positive integern, thenf (x) =nxn our definition, we havef (x) = limh 0f(x+h) f(x)h= limh 0(x+h)n xnh= limh 0 nk=0n!k!(n k)!hkxn k xnh= limh 0 nk=1n!k!(n k)!hkxn kh= limh 0n k=1n!k!(n k)!hk 1xn k=nxn the derivative ofp(x) =anxn+an 1xn 1+ +a1x+a0as a function (Solution) We can use the fact that the derivative splits over sums and the power rule tofind thatp (x) =ddx(anxn) +ddx(an 1xn 1) + +ddx(a1x) +ddx(a0)=nanxn 1+ (n 1)an 1xn 2+ +