math 470 module 4 discussion.docx - 3.2.1 lim mx b =mc b...

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3.2.1 Given that lim x→ c ( mx + b ) = mc + b Suppose ε > 0 given such that |(mx+b)-(mc+b) |< ε |m(x-c) |< ε |m||x-c|< ε | x-c| < ε | m | Then, by definition, we have Given ε > 0 there exists δ (= ε | m | ¿ > 0 such that |(mx+ b) – (mc +b)| < ε whenever | x- c| < δ Hence lim x→ c ( mx + b ) = mc + b 3.2.2(2) To show lim x → 3 x 2 9 x 3 = 6 If we let f(x)= x 2 9 x 3 Now, for all ϵ > 0, δ such that |f(x)-6| ¿ ϵ whenever 0< |x-3|< δ Consider |f(x)-6| = | x 2 9 x 3 6 ¿ x 2 9 6 x + 18 x 3 ¿ =| x 2 6 x + 9 x 3 ¿ =| ( x 3 ) 2 x 3 ¿ |x-3| Now |f(x)-6|< ϵ whenever 0 <|x-3|< δ
Then ϵ = δ 3.2.5 Given: J I R be open intervals. c J and f : I { c ) →R is a function. Prove: lim x→ c f ( x ) exists iff lim x→ c f | J ( x ) exists and if these exists then they must be equal. Proof Assume that lim x→ c f ( x ) =>For all ϵ > 0, δ such that for all x I { c } If 0 < |x-c| < δ then |f(x) – f ( c) | < Since J I , Now we can find a δ such that for all ϵ > 0, If 0 < | x – c | < δ , then |f(x) – f (c) | < lim x→ c f ( x ) exists. So, let’s assume that lim x→ c f | J ( x )

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