Continuity at a point of a piecewise function
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5
down vote
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Let
$
f(x) = left{
begin{array}{lr}
frac{x}{3} & x ~textrm{is rational},\
x & x ~textrm{is irrational}.
end{array}
right.
$
Show that $f$ is continuous at $a in mathbb{R}$ if and only if $a=0$.
My initial approach is to use the sequential criterion with the use of density of rational numbers but I wasn't successful. Any help is much appreciated.
calculus real-analysis analysis functions continuity
edited Nov 15 at 0:05
Rebellos
11.7k21040
add a comment |
up vote
5
down vote
favorite
Let
$
f(x) = left{
begin{array}{lr}
frac{x}{3} & x ~textrm{is rational},\
x & x ~textrm{is irrational}.
end{array}
right.
$
Show that $f$ is continuous at $a in mathbb{R}$ if and only if $a=0$.
My initial approach is to use the sequential criterion with the use of density of rational numbers but I wasn't successful. Any help is much appreciated.
calculus real-analysis analysis functions continuity
edited Nov 15 at 0:05
Rebellos
11.7k21040
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let
$
f(x) = left{
begin{array}{lr}
frac{x}{3} & x ~textrm{is rational},\
x & x ~textrm{is irrational}.
end{array}
right.
$
Show that $f$ is continuous at $a in mathbb{R}$ if and only if $a=0$.
My initial approach is to use the sequential criterion with the use of density of rational numbers but I wasn't successful. Any help is much appreciated.
calculus real-analysis analysis functions continuity
edited Nov 15 at 0:05
Rebellos
11.7k21040
Let
$
f(x) = left{
begin{array}{lr}
frac{x}{3} & x ~textrm{is rational},\
x & x ~textrm{is irrational}.
end{array}
right.
$
Show that $f$ is continuous at $a in mathbb{R}$ if and only if $a=0$.
My initial approach is to use the sequential criterion with the use of density of rational numbers but I wasn't successful. Any help is much appreciated.
calculus real-analysis analysis functions continuity
calculus real-analysis analysis functions continuity
edited Nov 15 at 0:05
Rebellos
11.7k21040
edited Nov 15 at 0:05
Rebellos
11.7k21040
edited Nov 15 at 0:05
Rebellos
11.7k21040
edited Nov 15 at 0:05
Rebellos
11.7k21040
edited Nov 15 at 0:05
Rebellos
11.7k21040
11.7k21040
asked Nov 14 at 23:50
user615914
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1 Answer
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2
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Let $x in mathbb R$, $x$ constant. If $(rho_n)$ be a sequence of rational numbers with $$lim rho_n = x Rightarrow lim f(rho_n) = lim rho_n/3 = x/3$$
If $(a_n)$ is a sequence of irrational numbers with
$$lim a_n = x Rightarrow lim f(a_n) = lim a_n = x$$
then, it is
$$lim f(rho_n)neqlim f(a_n) Leftrightarrow x neq 0$$
Thus, $f$ is not continuous at $mathbb R setminus {0}$.
We will now show that $f$ is continuous at $0$. It is $f(0) =0$. For all $|x| < 1$, we have :
$$|f(x) - f(0)| = |f(x)| leq max{|x|,|x|/3} = |x|$$
thus $lim_{xto 0} f(x) = f(0)$ which means that $f$ is continuous at $0$.
answered Nov 15 at 0:05
Rebellos
11.7k21040
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $x in mathbb R$, $x$ constant. If $(rho_n)$ be a sequence of rational numbers with $$lim rho_n = x Rightarrow lim f(rho_n) = lim rho_n/3 = x/3$$
If $(a_n)$ is a sequence of irrational numbers with
$$lim a_n = x Rightarrow lim f(a_n) = lim a_n = x$$
then, it is
$$lim f(rho_n)neqlim f(a_n) Leftrightarrow x neq 0$$
Thus, $f$ is not continuous at $mathbb R setminus {0}$.
We will now show that $f$ is continuous at $0$. It is $f(0) =0$. For all $|x| < 1$, we have :
$$|f(x) - f(0)| = |f(x)| leq max{|x|,|x|/3} = |x|$$
thus $lim_{xto 0} f(x) = f(0)$ which means that $f$ is continuous at $0$.
answered Nov 15 at 0:05
Rebellos
11.7k21040
add a comment |
up vote
2
down vote
Let $x in mathbb R$, $x$ constant. If $(rho_n)$ be a sequence of rational numbers with $$lim rho_n = x Rightarrow lim f(rho_n) = lim rho_n/3 = x/3$$
If $(a_n)$ is a sequence of irrational numbers with
$$lim a_n = x Rightarrow lim f(a_n) = lim a_n = x$$
then, it is
$$lim f(rho_n)neqlim f(a_n) Leftrightarrow x neq 0$$
Thus, $f$ is not continuous at $mathbb R setminus {0}$.
We will now show that $f$ is continuous at $0$. It is $f(0) =0$. For all $|x| < 1$, we have :
$$|f(x) - f(0)| = |f(x)| leq max{|x|,|x|/3} = |x|$$
thus $lim_{xto 0} f(x) = f(0)$ which means that $f$ is continuous at $0$.
answered Nov 15 at 0:05
Rebellos
11.7k21040
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $x in mathbb R$, $x$ constant. If $(rho_n)$ be a sequence of rational numbers with $$lim rho_n = x Rightarrow lim f(rho_n) = lim rho_n/3 = x/3$$
If $(a_n)$ is a sequence of irrational numbers with
$$lim a_n = x Rightarrow lim f(a_n) = lim a_n = x$$
then, it is
$$lim f(rho_n)neqlim f(a_n) Leftrightarrow x neq 0$$
Thus, $f$ is not continuous at $mathbb R setminus {0}$.
We will now show that $f$ is continuous at $0$. It is $f(0) =0$. For all $|x| < 1$, we have :
$$|f(x) - f(0)| = |f(x)| leq max{|x|,|x|/3} = |x|$$
thus $lim_{xto 0} f(x) = f(0)$ which means that $f$ is continuous at $0$.
answered Nov 15 at 0:05
Rebellos
11.7k21040
Let $x in mathbb R$, $x$ constant. If $(rho_n)$ be a sequence of rational numbers with $$lim rho_n = x Rightarrow lim f(rho_n) = lim rho_n/3 = x/3$$
If $(a_n)$ is a sequence of irrational numbers with
$$lim a_n = x Rightarrow lim f(a_n) = lim a_n = x$$
then, it is
$$lim f(rho_n)neqlim f(a_n) Leftrightarrow x neq 0$$
Thus, $f$ is not continuous at $mathbb R setminus {0}$.
We will now show that $f$ is continuous at $0$. It is $f(0) =0$. For all $|x| < 1$, we have :
$$|f(x) - f(0)| = |f(x)| leq max{|x|,|x|/3} = |x|$$
thus $lim_{xto 0} f(x) = f(0)$ which means that $f$ is continuous at $0$.
answered Nov 15 at 0:05
Rebellos
11.7k21040
answered Nov 15 at 0:05
Rebellos
11.7k21040
answered Nov 15 at 0:05
Rebellos
11.7k21040
answered Nov 15 at 0:05
Rebellos
11.7k21040
11.7k21040
add a comment |
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