Asymptotes and uniform continuity Let f:R->R be a continuous function with oblique asymptotes for x->+- infinity. Prove that the function is uniformly continuous. Proof: Let in>0. Since the function has obliquous asymptotes then ∃M>0:|f(x)−(ax+b)|<inAAx>M

Antwan Perez 2022-10-25 Answered
Asymptotes and uniform continuity
Let f : R R be a continuous function with oblique asymptotes for x ± . Prove that the function is uniformly continuous.
Proof: Let . Since the function has obliquous asymptotes then
M > 0 : | f ( x ) ( a x + b ) | < ϵ x > M
and
m > 0 : | f ( x ) ( c x + d ) | < ϵ x < m
Now, in the interval the function is uniformly continuous because of the Heine-Cantor theorem. We deal with the case + since the other one is formally identical. We know that lines are uniformly continuous. So, using the that certainly exists for the line we have:
| f ( x ) f ( x 0 ) | = | f ( x ) ( a x 0 + b ) + ( a x 0 + b ) f ( x 0 ) | | f ( x ) ( a x 0 + b ) | + | ( a x 0 + b ) f ( x 0 ) | < | f ( x ) ( a x 0 + b ) | + ϵ
But
| f ( x ) ( a x 0 + b ) | | f ( x ) ( a x + b ) | + | ( a x + b ) ( a x 0 + b ) | < 2 ϵ
So we are finished because if a continuous function is uniformly continuous in and , then it is uniformly continuous in .
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Answers (1)

Tania Alvarado
Answered 2022-10-26 Author has 15 answers
Your proof looks good to me, if I understand it right. I have one remark:
The core idea of the proof is to show that the function is uniformly continuous on the sets ( , m 1 ], [ m 1 , M + 1 ], and [ M + 1 , ).
Since is continuous at M + 1 and m 1, this is sufficient for it to be uniformly continuous everywhere. (See Hagen von Eitzen comments.) Alternatively, a perhaps easier way is to consider the intervals ( , m 1 2 ], [ m 1 , M + 1 ], and [ M + 1 2 , ).
Since these intervals actually overlap, you can show more easily that the must be uniformly continuous on the union if it is uniformly continuous on the individual sets (just take to be the minimum of all the deltas for each set and the length of the overlap).
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