32 Suppose c and of limits: a) are real num bers. Prove each the following using the E -S definition of lim b)文C X =c c) lim Ix) = lc[ Hint: first show V= d) Qim メーC a) m Vx VC for c>o (Hint: first show I-vel: x-c| %3D 36 6. FUNCTIONS AND CONTINUITY Definition. Let E CRand f a real-valued function with domain E. Let c be a limit point of E. Then f has a limit of L at c, denoted lim f(x) = L or f(x) → L as x c, if for all e > 0, there %3D exists a d(c, e) > 0 such that for all x E E with 0< x - c < 8, \f (x) – LEE. Example. lim(5.x + 7) = 22. L. Proof. Let e > 0. Define & = Then for all x ER with 0 < |x – 3| < 8, L %3D |(5x +7)- 22| = [5x - 15| 3D5(x-3) c-S 5€ < 58 = = E. So 5x +7 22 as x 3. Example. lim(x² + 2x – 7) = 41. %3D Proof. Let e > 0. Define & = min {1,. Then for all r € R with 0 < |x – 6| < 8, %3D -48 = |2 - 6|| + 8| ad 8,-0 SEDa< 15|x – 6| |(x2 +2x - 7)- 41| = |x² + 2x – 48|| %3D %3D need a bound 1x-6/<5 十14 IX49] <15 15€ < 158 < 15 1585 € = e. とx18く4tS S178지7회 So r2 + 2x-7 41 as r6. S-mingi, Example. lim(r² – 7x) = -10. Proof. Let e> 0. Define &= min {1,. Then for all r eR with 0< |r - 2| < 8, |(2² - 7x) + 10| = |r- 2||a- 5| -3-3-3

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