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40 CHAPTER 3. CONTINUOUS FUNCTIONS OF ONE VARIABLE

This proves the first part of 2.) To obtain the second part, let δ 1 be as described above andlet δ 0 > 0 be such that for |x−y|< δ 0,

|g(x)−g(y)|< |g(x)|/2

and so by the triangle inequality,

−|g(x)|/2≤ |g(y)|− |g(x)| ≤ |g(x)|/2

which implies |g(y)| ≥ |g(x)|/2, and |g(y)|< 3 |g(x)|/2.Then if |x−y|< min(δ 0,δ 1) ,∣∣∣∣ f (x)

g(x)− f (y)

g(y)

∣∣∣∣= ∣∣∣∣ f (x)g(y)− f (y)g(x)g(x)g(y)

∣∣∣∣≤ | f (x)g(y)− f (y)g(x)|(

|g(x)|22

)=

2 | f (x)g(y)− f (y)g(x)||g(x)|2

≤ 2

|g(x)|2[| f (x)g(y)− f (y)g(y)+ f (y)g(y)− f (y)g(x)|]

≤ 2

|g(x)|2[|g(y)| | f (x)− f (y)|+ | f (y)| |g(y)−g(x)|]

≤ 2

|g(x)|2

[32|g(x)| | f (x)− f (y)|+(1+ | f (x)|) |g(y)−g(x)|

]≤ 2

|g(x)|2(1+2 | f (x)|+2 |g(x)|) [| f (x)− f (y)|+ |g(y)−g(x)|]

≡M [| f (x)− f (y)|+ |g(y)−g(x)|]

where M is defined by

M ≡ 2

|g(x)|2(1+2 | f (x)|+2 |g(x)|)

Now let δ 2 be such that if |x−y|< δ 2, then

| f (x)− f (y)|< ε

2M−1

and let δ 3 be such that if |x−y|< δ 3, then

|g(y)−g(x)|< ε

2M−1.

Then if 0 < δ ≤min(δ 0,δ 1,δ 2,δ 3) , and |x−y|< δ , everything holds and∣∣∣∣ f (x)g(x)

− f (y)g(y)

∣∣∣∣≤M [| f (x)− f (y)|+ |g(y)−g(x)|]

40 CHAPTER 3. CONTINUOUS FUNCTIONS OF ONE VARIABLEThis proves the first part of 2.) To obtain the second part, let 6; be as described above andlet 59 > 0 be such that for |x—y| < 60,lg (x) —(y)1 < |g (x)|/2and so by the triangle inequality,—|g(*)|/2<|sQ)|-Is@1< |g @)|/2which implies |g (y)| > |g (x)| /2, and |g (y)| < 3|g(x)| /2.Then if |x—y| < min (60,61),f(x) £m) — |e) Foe)(x) g(y)g(x) 8) g(xaly< f@)s0)—f)8 0)— (et)_ 2\f as) -fO) 8)Ig (x)?< ae Ife) -FO)80)+f£0)80)-FO) 8 @)]* oR (le OIL @)—FO+FLF OI) —8 Wl<p [sls MIPe)—-FOI HG +E) 80) —e)<p t+ 2U I +2le ODI) ~F0)1+ 186) -8C0]=M(If(x)-fFM1+|80) —s @)lwhere M is defined by2M= 5 (1+2|f(a)| +218 @)/)Is (x)|Now let 62 be such that if |x—y| < 62, thenEE. _If) - FO) < 5Mand let 63 be such that if |x—y| < 63, then€Igy) -8(@)| < 5M".Then if 0 < 6 < min(60, 6), 62,63), and |x—y| < 6, everything holds andF8) _ £0) < wpe) -Pe LON mile f(y) +le(v) — g(x)