The proof that if f(x) is continuous at x=c, then 1f(x) is continuous at x=c...

lugreget9

lugreget9

Answered

2022-01-14

The proof that if f(x) is continuous at x=c, then 1f(x) is continuous at x=c
If f(x) is continuous at x=c, f(x)0 then 1f(x) is continuous at x=c.
My proof: for any ϵ>0, we need to show there exists δ>0, such that whenever |xc|<δ, we have |1f(x)1f(c)|<ϵ.
|1f(x)1f(c)|=|f(x)f(c)f(x)f(c)|
=|f(x)f(c)|1|f(x)f(c)|
=|f(x)f(c)|1|f(x)|1|f(c)|
Since f is continuous at c, thus |f(x)f(c)|<ϵ. I need to find an upper bound of |f(x)|.
Because |f(x)f(c)|<ϵ, we then have
|f(c)|=|f(c)f(x)+f(x)||f(c)f(x)|+|f(x)|.
<ϵ+|f(x)|
Thus, we find |f(x)|>|f(c)|ϵ. Equivalently, 1|f(x)|<1|f(c)|ϵ. (Not rigorous here, since |f(c)|ϵ could be negative, how to remedy?)
Finally, (ignore not rigorous part)
|1f(x)1f(c)|<ϵ|f(c)|ϵ1|f(c

Answer & Explanation

Bob Huerta

Bob Huerta

Expert

2022-01-15Added 41 answers

You can start by taking δ0 such that
|xc|<δ0|f(x)f(c)|<|f(c)|2.
Therefore |xc|<δ0|f(x)|>f(c)2
and so |xc|<δ0|1f(x)1f(c)|<2|f(x)f(c)|f2(c)
So, if you take δ1>0 such that
|xc|<δ1|f(x)f(c)|<ξf2(c)2,
then |xc|<min{δ0,δ1}|1f(x)1f(c)|<ξ.
Samantha Brown

Samantha Brown

Expert

2022-01-16Added 35 answers

Suppose we want to find a lower bound of f(x) in a neighbourhood of a where f(a)0. Since f is continuous we can choose |f(x)f(a)|<ξ in a neighbourhood of a. The trick here is choosing ξ=|f(a)|2>0. We then get |f(x)f(a)|<|f(a)|2. We also have |f(a)||f(x)||f(x)f(a)| by the triangle inequality, and we get |f(a)||f(x)|<|f(a)|2. It then follows that |f(a)|2<|f(x)|.

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