Prove the following limit. lim5x−4=6 x→2 SOLUTION 1. Preliminary analysis of the problem guessing a value for δ Let € be a given positive number. We want to find a number δ such that if 0∣&lt;x−2∣&lt;δ then∣(5x−4)−6∣&lt;ϵ. But ∣(5x−4)−6∣=∣5x−10∣=5||. Therefore, we want δ if 0&lt;∣x−2∣&lt;δ then5∣∣ &lt;ϵ such that that is, if 0&lt;∣x−2∣&lt;δ then∣∣&lt;ϵϵ5 This suggests that we should choose δ=ϵ/5. 2. Proof (showing that δ works). Given ϵ&gt;0, choose δ= ϵ/5.If 0&lt;∣∣&lt;δ, then ∣(5x−4)−6∣=∣∣ =5∣∣&lt;5δ =5() =ϵ Thus if 0&lt;∣x−2∣&lt;δ then∣(5x−4)−6∣&lt;ϵ. Therefore, by the definition of a limit lim5x−4=6 x→2

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Prove the following limit. lim5x−4=6 x→2 SOLUTION 1. Preliminary analysis of the problem guessing a value for δ Let € be a given positive number. We want to find a number δ such that if 0∣&amp;lt;x−2∣&amp;lt;δ then∣(5x−4)−6∣&amp;lt;ϵ. But ∣(5x−4)−6∣=∣5x−10∣=5||. Therefore, we want δ if 0&amp;lt;∣x−2∣&amp;lt;δ then5∣∣ &amp;lt;ϵ such that that is, if 0&amp;lt;∣x−2∣&amp;lt;δ then∣∣&amp;lt;ϵϵ5 This suggests that we should choose δ=ϵ/5. 2. Proof (showing that δ works). Given ϵ&amp;gt;0, choose δ=  ϵ/5.If 0&amp;lt;∣∣&amp;lt;δ, then ∣(5x−4)−6∣=∣∣ =5∣∣&amp;lt;5δ =5() =ϵ Thus if 0&amp;lt;∣x−2∣&amp;lt;δ then∣(5x−4)−6∣&amp;lt;ϵ. Therefore, by the definition of a limit lim5x−4=6 x→2

Roger Noah · Accepted Answer

Given, lim5x−4=6 x→2 Let ϵ be a given positive number. We want to find a number δ such that, if 0&amp;lt;∣x—2∣&amp;lt;δ then ∣(5x−4)—6∣&amp;lt;ϵ. But ∣(5x—4)—6∣=∣5x—10∣ =∣5(x−2)∣ =5∣x−2∣ Therefore we want δ such that if 0&amp;lt;∣x—2∣&amp;lt;δ then 5∣x—2∣&amp;lt;ϵ. That is, if 0&amp;lt;∣x—2∣&amp;lt;δ then∣x—2∣&amp;lt;ϵ5. This suggests that we should choose δ=ϵ5 Step 2 Given ϵ&amp;gt;0, choose δ=ϵ5. if 0&amp;lt;∣x—2∣&amp;lt;δ, then ∣(5x—4)−6∣=∣5x—10∣ =5∣x−2∣&amp;lt;5δ =5(∣x−2∣&amp;lt;δ) =5(ϵ5) =ϵ Thus, if 0&amp;lt;∣x—2∣&amp;lt;δ then∣(5x−4)−6∣&amp;lt;ϵ Therefore by the definition of a limit, lim(5x−4)=6 x→2  Step 2 Given ϵ&amp;gt;0, choose δ=ϵ5. if 0&amp;lt;∣x—2∣&amp;lt;δ, then ∣(5x—4)−6∣=∣5x—10∣ =5∣x−2∣&amp;lt;5δ =5(∣x−2∣&amp;lt;δ) =5(ϵ5) =ϵ Thus, if 0&amp;lt;∣x—2∣&amp;lt;δ then∣(5x−4)−6∣&amp;lt;ϵ. Therefore by the definition of a limit, lim(5x−4)=6 x→2

Prove the following limit.\lim 5x-4=6x\rightarrow2Z

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