# Student 3 (2).docx - Nicholas Wholean Student 3 3.2.1 Let...

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Nicholas Wholean Student 3 3.2.1 Let m,b,c . Using only the definition of limits, prove that lim x→ c ( mx + b ) = mc + b We can look at f(x)=mx+b and know that f: and c is the limit of the domain First, we must prove that for every ε>0 there is exists an δ>0 such that l f(x)-(mc+b) l < ε for l x-c l < δ, x Let’s look at l f(x)-(mc+b) l= l mx+b-mc-b l =l mx-mc l I f(x)-(mc+b) l= lml lx-cl Then we can move on too, l (mx+b)-(mc+b) l<ε lml lx-cl <ε lx-cl< ε m Next we can look at δ = ε m Then l (mx+b)-(mc+b) l <ε whenever l x-c l<δ Therefore if ε >0 there exists a δ>0 such that l (mx+b)-(mc+b) l < ε whenever l x-c l<δ Or, in other words lim x→ c ( mx + b ) = mc + b 3.2.3 Let I ⊆ℝ be an open interval, let c I, let f:I-{c}→ be an function and let L ∈ℝ . Using only the definition of limits, prove that lim x→ c f ( x ) = L , if and only if lim x→ c [ f ( x ) L ] = 0 lim x→ c f ( x ) = L , for every ε>0 there is a δ>0 such that l f(x)-L l<ε wherever 0<l x-0 l <δ lim x→ c f ( x ) L = 0 , we know for some ε>0 and δ>0 l f(x)-L)-0 l <ε whenever 0<l x-c l<δ, then l f(x)-L l <ε, then lim x →C f ( x ) =