MTH 251 – Differential Calculus Chapter 2 – Limits and Continuity(Two Sided) LimitsOne Sided Limits(One Sided) Right-Hand Limits(One Sided) Left-Hand LimitsExamples …When Left & Right Limits are EqualA special & important limit.Slide 9Slide 10Variations ofMore examples …One more example …MTH 251 – Differential CalculusChapter 2 – Limits and ContinuitySection 2.4One-Sided LimitsCopyright © 2010 by Ron Wallace, all rights reserved.(Two Sided) Limits•For to exist, the function MUST be defined for SOME (maybe small) open interval around c (excluding c). •Review formal definition:lim ( )x cf x�lim ( ) if,for all 0 there is a 0 such thatif 0 , then ( )x cf x Lx c f x Le dd e�=> >< - < - <One Sided Limits•There are some cases where the function is defined on only one side of c or is defined differently on each side of c.Examples …21lim 1xx�-|x| can only be smaller than 1.222 1, 2lim ( ) where ( )3, 2xx xf x f xx x�+ <�=�- ��That is … these two limits do not exist!•Only values to the right of c (i.e. on the positive side or greater than c) need to be considered.•Definition? Essentially the same, except the interval for  is changed to …(One Sided) Right-Hand Limitsc x c d< < +lim ( )x cf x+�L + L - Lcc + !•Only values to the left of c (i.e. on the negative side or less than c) need to be considered.•Definition? Essentially the same, except the interval for  is changed to …(One Sided) Left-Hand Limitsc x cd- < <lim ( )x cf x-�L + L - Lcc - !Examples …•Determine the following limits for the graph …2lim ( )xf x-� 2lim ( )xf x+�3lim ( )xf x-� 3lim ( )xf x+�3lim ( )xf x+�-4lim ( )xf x-�4lim ( )xf x+�0lim ( )xf x�When Left & Right Limits are Equal… if and only if …lim ( )x cf x L�=lim ( ) lim ( )x c x cf x L f x- +� �= =NOTE: Sometimes, in order to determine a limit,you must consider the left and right hand limits individually.A special & important limit.0sinlimxxx+�ABCD1xA = (0, 0)B = (1, 0)C = (cos x, sin x)D = (1, tan x)area small < area sector < area big D Dsin1 1 12 2 2 cossinxxx x< <1sin cos1xx x< <sincos 1xxx < <sin0 0 0lim cos lim lim 1xxx x xx+ + +� � �< <1=First consider the right hand limit.A special & important limit.0sinlimxxx-�area small < area sector < area big D Dsin1 1 12 2 2 cossinxxx x- <- <-1cos sin1xx x< <sin1 cosxxx< <sin0 0 0lim 1 lim lim cosxxx x xx- - -� � �< <1=BxA = (0, 0)B = (1, 0)C = (cos x, sin x)D = (1, tan x)DCA 1Next consider the left hand limit.A special & important limit.0sinlimxxx-�1=0sinlimxxx+�1=Since …… it can be concluded that …0sinlimxxx�1=NOTE: Remember this result, it will be needed later.Variations of 0sinlimxxx�1=0sin 5lim5xxx�1=0sinlimuuu�=Let u = 5x0sin 5lim3xxx�05 sin 5lim3 5xxx�=53=More examples …64sin 2lim5xxxp�04sin 2lim5xxx�0cos 1limxxx�-0sin 3limsin 7xxx�One more example …20coslimsin 3xx x