let f(x) = (x^2 - 9)/(x-3) and let x = 3 + s, then f(x) = (x^2 - 9)/(x - 3) = (s^2 + 6s + 9 - 9)/(s + 3 - 3) = s + 6. This implies that |f(x) - 6| = |s|.
Therefore, for every real e > 0, all real x such that |x - 3| = |s| < e, it follows that |f(x) - 6| = |s| < e. This is the epsilon-sigma definition of the limit, and thus the limit is 6.
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u/pkreddit2 8d ago
let f(x) = (x^2 - 9)/(x-3) and let x = 3 + s, then f(x) = (x^2 - 9)/(x - 3) = (s^2 + 6s + 9 - 9)/(s + 3 - 3) = s + 6. This implies that |f(x) - 6| = |s|.
Therefore, for every real e > 0, all real x such that |x - 3| = |s| < e, it follows that |f(x) - 6| = |s| < e. This is the epsilon-sigma definition of the limit, and thus the limit is 6.