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.

It seems like there are some missing parts in your question, making it difficult to understand what you're asking. However, I'll try to explain based on what I can gather.

The statement "if 0 < |x − 5| < 𝛿 then 5 < 𝜀" is a way of expressing the definition of a limit in calculus. It's saying that for any positive number 𝜀, no matter how small, we can find a positive number 𝛿 such that if the distance between x and 5 is less than 𝛿 (but not zero), then the distance between 5x - 7 and 18 is less than 𝜀.

The choice of 𝛿 = 𝜀/5 is a way to ensure this condition is met. The proof then shows that if 0 < |x − 5| < 𝛿, then |(5x − 7) − 18| < 𝜀, confirming that the choice of 𝛿 works.

Finally, the statement "lim x → 5 5x − 7 = 18" is the conclusion that the limit as x approaches 5 of the function 5x - 7 is 18, which is what the initial conditions and proof were leading up to.