Calculus Some Examples of Continuous Functions September 20, 2018
Example. Letk ∈N={1,2, . . .}, n = 2k+ 1 and let
f(x) = x1/n forx∈R= (−∞,∞).
Show thatf is continuous everywhere in R.
Proof. Case 1: a6= 0 =⇒ |a|>0.
For each >0, since
|f(x)−f(a)|=|x1/n−a1/n|= |x−a|
|xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n|, choose
δ= min{|a|,|a|n−1/n} such that if |x−a|< δ then
|xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n|
= |x|n−1/n+|x|n−2/n|a|1/n+· · ·+|x|1/n|a|n−2/n+|a|n−1/n
≥ |a|n−1/n
and
|f(x)−f(a)|= |x−a|
|xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n|< δ
|a|n−1/n ≤ |a|n−1/n
|a|n−1/n =. This proves thatf is continuous at eacha 6= 0.
Case 2: a= 0.
For each >0, choose
δ =n such that if |x−0|< δ then
|f(x)−f(0)|=|x1/n|=|x|1/n =|x−0|1/n<δ1/n= n1/n
=. This proves thatf is continuous at a= 0.
Example. Letk ∈N={1,2, . . .}, n = 2k and let
f(x) = x1/n for x∈[0,∞).
Show thatf is continuous everywhere in [0,∞).
Proof. Case 1: a >0.
For each >0, since
|f(x)−f(a)|=|x1/n−a1/n|= |x−a|
|xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n|,
Calculus Some Examples of Continuous Functions (Continued) September 20, 2018 we choose
δ= min{a, an−1/n} such that if |x−a|< δ then
|xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n|
= xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n
≥ an−1/n
and
|f(x)−f(a)|= |x−a|
|xn−1/n+xn−2/na1/n+· · ·+x1/nan−2/n+an−1/n| < δ
an−1/n ≤ an−1/n an−1/n =. This proves thatf is continuous at eacha >0.
Case 2: a= 0.
For each >0, we choose
δ =n such that if |x−0|< δ then
|f(x)−f(0)|=|x1/n|=|x|1/n =|x−0|1/n < δ1/n= n1/n
=. This proves thatf is continuous at a= 0.
Example. Letf be continuous at a and f(a)6= 0. Show that 1
f(x) is continuous at a.
Proof.
Since f be continuous at a and f(a)6= 0 which implies that
|f(a)|>0, f(a)−|f(a)|
2 6= 0 and f(a) + |f(a)|
2 6= 0 and there exists a δ1 >0 such that if if|x−a|< δ1 then
|f(x)−f(a)|< |f(a)|
2
=⇒ −|f(a)|
2 < f(x)−f(a)< |f(a)|
2
=⇒ f(a)−|f(a)|
2 < f(x)< f(a) + |f(a)|
2
=⇒ |f(x)| ≥L= min{|f(a)− |f(a)|
2 |,|f(a) + |f(a)|
2 |}>0.
For each >0, since f be continuous ata, there exists and δ2 >0 such that if|x−a|< δ2 then
|f(x)−f(a)|< L|f(a)|.
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Calculus Some Examples of Continuous Functions (Continued) September 20, 2018 By setting
δ = min{δ1, δ2}, and if |x−a|< δ then
| 1
f(x) − 1
f(a)|= |f(x)−f(a)|
|f(x)||f(a)| < L|f(a)|
|f(x)||f(a)| ≤. This proves that 1
f(x) is continuous at a.
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