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Find the largest number 𝛿 such that if |x − 1| < 𝛿, then |5x − 5| < 𝜀, where 𝜀 = 1.𝛿 ≤ Repeat and determine 𝛿 with 𝜀 = 0.1.𝛿 ≤

Question

Find the largest number 𝛿 such that if |x − 1| < 𝛿, then |5x − 5| < 𝜀, where 𝜀 = 1.𝛿 ≤ Repeat and determine 𝛿 with 𝜀 = 0.1.𝛿 ≤

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Solution

The problem is asking for a value of 𝛿 that satisfies the given conditions. This is a typical problem in calculus, specifically in the concept of limits.

Step 1: We start by writing down the given inequality |5x - 5| < 𝜀.

Step 2: We can simplify this inequality by factoring out 5 to get |5(x - 1)| < 𝜀.

Step 3: Since the absolute value of a product is the product of the absolute values, we can rewrite this as 5|x - 1| < 𝜀.

Step 4: Now we want to isolate |x - 1|, so we divide both sides of the inequality by 5 to get |x - 1| < 𝜀/5.

Step 5: We know that 𝛿 is the value such that |x - 1| < 𝛿, so we can set 𝜀/5 = 𝛿.

Step 6: If 𝜀 = 1, then 𝛿 = 1/5 = 0.2.

Step 7: If 𝜀 = 0.1, then 𝛿 = 0.1/5 = 0.02.

So, the largest number 𝛿 that satisfies the given conditions is 0.2 when 𝜀 = 1, and 0.02 when 𝜀 = 0.1.

This problem has been solved

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if      0 < |x − 5| < 𝛿      then      5  <  𝜀that is,          if      0 < |x − 5| < 𝛿      then       <  𝜀5.This suggests that we should choose 𝛿 = 𝜀/5.2. Proof (showing that 𝛿 works). Given 𝜀 > 0, choose 𝛿 = 𝜀/5. If 0 <   < 𝛿, then|(5x − 7) − 18|  =   =  5  < 5𝛿 =  5   =  𝜀.Thusif    0 < |x − 5| < 𝛿    then    |(5x − 7) − 18| < 𝜀.Therefore, by the definition of a limitlim x → 5 5x − 7 = 18.

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