Trigonometric Limits - California State University, …

1 Trigonometric Limitsmore examples of Limits Typeset by FoilTEX 1 Substitution Theorem forTrigonometric Functionslaws for evaluating Limits Typeset by FoilTEX 2 Theorem each pointcin function sdomain:limx csinx= sinc,limx ccosx= cosc,limx ctanx= tanc,limx ccotx= cotc,limx ccscx= cscc,limx csecx= each pointcin function sdomain:limx csinx= sinc,limx ccosx= cosc,limx ctanx= tanc,limx ccotx= cotc,limx ccscx= cscc,limx csecx= first thatlimx 0sinx= 0,limx 0cosx= 1. Typeset by FoilTEX 3Is it obvious?limx 0sinx= 0,limx 0cosx= (x)y=cos(x)Is it obvious?limx 0sinx= 0,limx 0cosx= (x)y=cos(x)No. The picture is not it obvious?limx 0sinx= 0,limx 0cosx= (x)y=cos(x)No. The picture is not definitions ofsin(x)andcos(x). Typeset by FoilTEX 4 Use The One-Sided Squeeze (x) g(x) h(x)nearcandlimx c+f(x) =limx c+h(x) =Lright,thenlimx c+g(x) =LrightcLy=h(x)y=f(x)y=g(x)Also, iflimx c f(x) = limx c h(x) =Lleft,thenlimx c g(x) =LleftcLy=h(x)y=f(x)y=g(x) Typeset by FoilTEX 5 Use The One-Sided cg(x) =L limx c g(x) = limx c+g(x) =LAnd other Limits c[f(x) +g(x)] = limx cf(x) + limx cg(x),limx c[f(x)g(x)] = [limx cf(x)][limx cg(x)], Typeset by FoilTEX 6An estimate from geometry:0<AB<AC<arcACOABCsin(t)tcos(t)2 1-cos(t)%&%&&&&&t<0t>01or,0<sin(t)< 2 1 cos(t)<tAn estimate from geometry.

2 0<AB<AC<arcACOABCsin(t)tcos(t)2 1-cos(t)%&%&&&&&t<0t>01or,0<sin(t)< 2 1 cos(t)<tBythe Right-Sided Squeeze Theoremlimx 0+sin(x) = 0,limx 0+(1 cos(x)) = 0, Typeset by FoilTEX 7 Similarly,limx 0 sin(x) = 0,limx 0 (1 cos(x)) = ,limx 0 sin(x) = 0,limx 0 (1 cos(x)) = left and the right Limits are equal, thuslimx 0sin(x) = 0,limx 0(1 cos(x)) = 0 Similarly,limx 0 sin(x) = 0,limx 0 (1 cos(x)) = left and the right Limits are equal, thuslimx 0sin(x) = 0,limx 0(1 cos(x)) = 0or,limx 0sin(x) = 0,limx 0cos(x) = 1. Typeset by FoilTEX 8 EXAMPLES Typeset by FoilTEX 9 EXAMPLE limitlim /4 tan( )Since = /4is in the domain of the function tan( )EXAMPLE limitlim /4 tan( )Since = /4is in the domain of the function tan( )we use Substitution Theorem to substitute /4for in the limit expression:lim /4 tan = 4tan( 4)= 4 1 = 4.

3 Typeset by FoilTEX 10 EXAMPLE limitlim /2cos2( )1 sin( ).Since at = /2the denominator ofcos2( )/(1 sin( ))turns to zero, we can not substitute /2for limitlim /2cos2( )1 sin( ).Since at = /2the denominator ofcos2( )/(1 sin( ))turns to zero, we can not substitute /2for , we rewrite the expressionusingsin2( ) + cos2( ) = 1:lim /21 sin2( )1 sin( )= lim /2(1 sin( ))(1 + sin( ))(1 sin( )) Typeset by FoilTEX 11 Finally,lim /2(1 sin( ))(1 + sin( ))(1 sin( ))Finally,lim /2(1 sin( ))(1 + sin( ))(1 sin( ))= lim /2(1 sin( )1 sin( ))lim /2(1 + sin( ))Finally,lim /2(1 sin( ))(1 + sin( ))(1 sin( ))= lim /2(1 sin( )1 sin( ))lim /2(1 + sin( ))= 1 (1 + sin( /2)) = 2. Typeset by FoilTEX 12 Special Trigonometric Limitssin(x)/x ?asx 0 Typeset by FoilTEX 13 Theorem 0sinxx= 01 cosxx= 0. Typeset by FoilTEX 14 Proof fact from geometry:(t>0)area(OAB) area(ODB) area(ODC)cos2(t)t/2 sin(t) cos(t)/2 bycos(t)t/2getcost sintt 1costRight-Sided Squeeze Theorem:limt 0+sintt= 1 Typeset by FoilTEX 15 The same inequality holds fort<0:cost sintt 1costLeft-Sided Squeeze Theorem.

4 Limt 0 sintt= 1 The left and the right Limits are equal, thus,limt 0sintt= 1 Typeset by FoilTEX 16 Proof multiplying numerator anddenominator with(1 + cosx)limx 01 cosxx= limx 0(1 cosx)x(1 + cosx)(1 + cosx)Proof multiplying numerator anddenominator with(1 + cosx)limx 01 cosxx= limx 0(1 cosx)x(1 + cosx)(1 + cosx)= limx 0(1 cos2x)x(1 + cosx)= limx 0sin2xx(1 + cosx)Proof multiplying numerator anddenominator with(1 + cosx)limx 01 cosxx= limx 0(1 cosx)x(1 + cosx)(1 + cosx)= limx 0(1 cos2x)x(1 + cosx)= limx 0sin2xx(1 + cosx)UsingB1write=[limx 0sinxx]limx 0[sinx]limx 0[1 + cosx]= 0. Typeset by FoilTEX 17 EXAMPLES Typeset by FoilTEX 18 EXAMPLE limitlimt 0tanttEXAMPLE limitlimt 0tanttRecallingtant= sint/cost,and usingB1:= limt 0sint(cost)tEXAMPLE limitlimt 0tanttRecallingtant= sint/cost,and usingB1:= limt 0sint(cost)t=[limt 0sintt]limt 01costEXAMPLE limitlimt 0tanttRecallingtant= sint/cost,and usingB1:= limt 0sint(cost)t=[limt 0sintt]limt 01cost= 1 11= 1 Typeset by FoilTEX 19 EXAMPLE limit (Can t useB1!)

5 :limt 0sin(3t)tEXAMPLE limit (Can t useB1!):limt 0sin(3t)tmultiply both numerator and denominator with3:= limt 03sin(3t)3tEXAMPLE limit (Can t useB1!):limt 0sin(3t)tmultiply both numerator and denominator with3:= limt 03sin(3t)3tNow,t 0as3t 0,so= lim3t 03sin(3t)3t= 3.(limx 0sinxx= 1)B1applies (with a substitutionx= 3t). Typeset by FoilTEX 20 EXAMPLE limitlimt 01 costsintEXAMPLE limitlimt 01 costsintDivide both numerator and denominator witht:= limt 01 costtsinttEXAMPLE limitlimt 01 costsintDivide both numerator and denominator witht:= limt 01 costtsinttUseB1andB2:=01= 0. Typeset by FoilTEX 21