Definition 2 For the right-hand limit we say that,limx→a+f(x)=Llimx→a+f(x)=L if for every number ε>0ε>0 there is some number δ>0δ>0 such that |f(x)−L|<εwhenever0<x−a<δ(or a<x<a+δ)|f(x)−L|<εwhenever0<x−a<δ(or a< x<a+δ) Definition 3 For the left-hand limit we say that,limx→a−f(x)=Llimx→a−f(x)=L if for every number ε>0ε>0 there is some number δ>0δ>0 such that |f(x)−L|<εwhenever−δ<x−a<0(or a−δ<x<a)|f(x)−L|<εwhenever−δ<x−a<0(or a−δ<x<a) Note that with both of these definitions there are two ways to deal with the restriction on xx and the one in parenthesis is probably the easier to use, although the main one given more closely matches the definition of the normal limit above. Let’s work a quick example of one of these, although as you’ll see they work in much the same manner as the normal limit problems do. Example 4 Use the definition of the limit to prove the following limit.limx→0+√ x =0limx→0+x=0 Hide Solution Let ε>0ε>0 be any number then we need to find a number δ>0δ>0 so that the following will be true. ∣∣√ x −0∣∣<εwhenever0<x−0<δ|x−0|<εwhenever0<x−0<δ Or upon a little simplification we need to show, √ x <εwhenever0<x<δx<εwhenever0<x<δ As with the previous problems let’s start with the left-hand inequality and see if we can’t use that to get a guess for δδ. The only simplification that we really need to do here is to square both sides. √ x <ε⇒x<ε2x<ε⇒x<ε2 So, it looks like we can choose δ=ε2δ=ε2. Let’s verify this. Let ε>0ε>0 be any number and chose δ=ε2δ=ε2. Next assume that 0<x<ε20<x<ε2. This gives, ∣∣√ x −0∣∣=√ x some quick simplification<√ ε2 use the assumption that x<ε2<εone final simplification|x−0|=xsome quick simplification<ε2use the assumption that x<ε2<εone final simplification We’ve now shown that, ∣∣√ x −0∣∣<εwhenever0<x−0<ε2|x−0|<εwhenever0<x−0<ε2 and so by the definition of the right-hand limit we have, limx→0+√ x =0limx→0+x=0 Let’s now move onto the definition of infinite limits. Here are the two definitions that we need to cover both possibilities, limits that are positive infinity and limits that are negative infinity.