Show if $f$ : differentiable at $c$, then $exists epsilon >0 $ s.t. $f$ is continuous on $N(c, epsilon)$
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Show if $f : mathbb{R} to mathbb{R}$ : differentiable at $c$, then $exists epsilon >0 $ s.t. $f$ is continuous on $N(c, epsilon)$.
My try
Since $f'(c) = lim_{h to 0}frac{f(c + h) - f(c)}{h}$, this means $lim_{h to 0} f(c + h) = f(c)$ (otherwise $f'(c)$ does not exists), which means $f$ is continuous on some neighbourhood of $c$.
But I'm stuck at showing this using $epsilon$-$delta$ method.
real-analysis
edited Nov 18 at 10:39
Robert Z
91.3k1058129
asked Nov 18 at 10:24
Moreblue
795216
add a comment |
up vote
1
down vote
favorite
Show if $f : mathbb{R} to mathbb{R}$ : differentiable at $c$, then $exists epsilon >0 $ s.t. $f$ is continuous on $N(c, epsilon)$.
My try
Since $f'(c) = lim_{h to 0}frac{f(c + h) - f(c)}{h}$, this means $lim_{h to 0} f(c + h) = f(c)$ (otherwise $f'(c)$ does not exists), which means $f$ is continuous on some neighbourhood of $c$.
But I'm stuck at showing this using $epsilon$-$delta$ method.
real-analysis
edited Nov 18 at 10:39
Robert Z
91.3k1058129
asked Nov 18 at 10:24
Moreblue
795216
Observe $f(c+h)-f(c)<hcdot f'(c)=epsilon$.
– Yadati Kiran
Nov 18 at 10:29
Well your statement is false. A typical counter-example is $f(x) =x^2,xinmathbb{Q},f(x)=-x^2,xinmathbb{R}setminus mathbb {Q} $. $f$ is continuous and differentiable at $0$ and at other points it is neither continuous nor differentiable.
– Paramanand Singh
Nov 18 at 14:47
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show if $f : mathbb{R} to mathbb{R}$ : differentiable at $c$, then $exists epsilon >0 $ s.t. $f$ is continuous on $N(c, epsilon)$.
My try
Since $f'(c) = lim_{h to 0}frac{f(c + h) - f(c)}{h}$, this means $lim_{h to 0} f(c + h) = f(c)$ (otherwise $f'(c)$ does not exists), which means $f$ is continuous on some neighbourhood of $c$.
But I'm stuck at showing this using $epsilon$-$delta$ method.
real-analysis
edited Nov 18 at 10:39
Robert Z
91.3k1058129
asked Nov 18 at 10:24
Moreblue
795216
Show if $f : mathbb{R} to mathbb{R}$ : differentiable at $c$, then $exists epsilon >0 $ s.t. $f$ is continuous on $N(c, epsilon)$.
My try
Since $f'(c) = lim_{h to 0}frac{f(c + h) - f(c)}{h}$, this means $lim_{h to 0} f(c + h) = f(c)$ (otherwise $f'(c)$ does not exists), which means $f$ is continuous on some neighbourhood of $c$.
But I'm stuck at showing this using $epsilon$-$delta$ method.
real-analysis
real-analysis
edited Nov 18 at 10:39
Robert Z
91.3k1058129
asked Nov 18 at 10:24
Moreblue
795216
edited Nov 18 at 10:39
Robert Z
91.3k1058129
asked Nov 18 at 10:24
Moreblue
795216
edited Nov 18 at 10:39
Robert Z
91.3k1058129
edited Nov 18 at 10:39
Robert Z
91.3k1058129
edited Nov 18 at 10:39
Robert Z
91.3k1058129
91.3k1058129
asked Nov 18 at 10:24
Moreblue
795216
asked Nov 18 at 10:24
Moreblue
795216
asked Nov 18 at 10:24
Moreblue
795216
795216
Observe $f(c+h)-f(c)<hcdot f'(c)=epsilon$.
– Yadati Kiran
Nov 18 at 10:29
Well your statement is false. A typical counter-example is $f(x) =x^2,xinmathbb{Q},f(x)=-x^2,xinmathbb{R}setminus mathbb {Q} $. $f$ is continuous and differentiable at $0$ and at other points it is neither continuous nor differentiable.
– Paramanand Singh
Nov 18 at 14:47
add a comment |
Observe $f(c+h)-f(c)<hcdot f'(c)=epsilon$.
– Yadati Kiran
Nov 18 at 10:29
Well your statement is false. A typical counter-example is $f(x) =x^2,xinmathbb{Q},f(x)=-x^2,xinmathbb{R}setminus mathbb {Q} $. $f$ is continuous and differentiable at $0$ and at other points it is neither continuous nor differentiable.
– Paramanand Singh
Nov 18 at 14:47
Observe $f(c+h)-f(c)<hcdot f'(c)=epsilon$.
– Yadati Kiran
Nov 18 at 10:29
Observe $f(c+h)-f(c)<hcdot f'(c)=epsilon$.
– Yadati Kiran
Nov 18 at 10:29
Well your statement is false. A typical counter-example is $f(x) =x^2,xinmathbb{Q},f(x)=-x^2,xinmathbb{R}setminus mathbb {Q} $. $f$ is continuous and differentiable at $0$ and at other points it is neither continuous nor differentiable.
– Paramanand Singh
Nov 18 at 14:47
Well your statement is false. A typical counter-example is $f(x) =x^2,xinmathbb{Q},f(x)=-x^2,xinmathbb{R}setminus mathbb {Q} $. $f$ is continuous and differentiable at $0$ and at other points it is neither continuous nor differentiable.
– Paramanand Singh
Nov 18 at 14:47
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
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Hint. Consider the function $f:mathbb{R}to mathbb{R}$, defined as
$$f(x)=begin{cases}
frac{1}{(2n+1)^2}&text{for $x=frac{1}{2n+1}$ and $nin mathbb{Z}$,}\
0 &text{otherwise.}
end{cases}$$
Is $f$ differentiable at $0$? Is $f$ continuous in $(-epsilon,epsilon)$ for some $epsilon>0$?
answered Nov 18 at 10:34
Robert Z
91.3k1058129
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Hint. Consider the function $f:mathbb{R}to mathbb{R}$, defined as
$$f(x)=begin{cases}
frac{1}{(2n+1)^2}&text{for $x=frac{1}{2n+1}$ and $nin mathbb{Z}$,}\
0 &text{otherwise.}
end{cases}$$
Is $f$ differentiable at $0$? Is $f$ continuous in $(-epsilon,epsilon)$ for some $epsilon>0$?
answered Nov 18 at 10:34
Robert Z
91.3k1058129
add a comment |
up vote
1
down vote
accepted
Hint. Consider the function $f:mathbb{R}to mathbb{R}$, defined as
$$f(x)=begin{cases}
frac{1}{(2n+1)^2}&text{for $x=frac{1}{2n+1}$ and $nin mathbb{Z}$,}\
0 &text{otherwise.}
end{cases}$$
Is $f$ differentiable at $0$? Is $f$ continuous in $(-epsilon,epsilon)$ for some $epsilon>0$?
answered Nov 18 at 10:34
Robert Z
91.3k1058129
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Hint. Consider the function $f:mathbb{R}to mathbb{R}$, defined as
$$f(x)=begin{cases}
frac{1}{(2n+1)^2}&text{for $x=frac{1}{2n+1}$ and $nin mathbb{Z}$,}\
0 &text{otherwise.}
end{cases}$$
Is $f$ differentiable at $0$? Is $f$ continuous in $(-epsilon,epsilon)$ for some $epsilon>0$?
answered Nov 18 at 10:34
Robert Z
91.3k1058129
Hint. Consider the function $f:mathbb{R}to mathbb{R}$, defined as
$$f(x)=begin{cases}
frac{1}{(2n+1)^2}&text{for $x=frac{1}{2n+1}$ and $nin mathbb{Z}$,}\
0 &text{otherwise.}
end{cases}$$
Is $f$ differentiable at $0$? Is $f$ continuous in $(-epsilon,epsilon)$ for some $epsilon>0$?
answered Nov 18 at 10:34
Robert Z
91.3k1058129
answered Nov 18 at 10:34
Robert Z
91.3k1058129
answered Nov 18 at 10:34
Robert Z
91.3k1058129
answered Nov 18 at 10:34
Robert Z
91.3k1058129
91.3k1058129
add a comment |
add a comment |
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Observe $f(c+h)-f(c)<hcdot f'(c)=epsilon$.
– Yadati Kiran
Nov 18 at 10:29
Well your statement is false. A typical counter-example is $f(x) =x^2,xinmathbb{Q},f(x)=-x^2,xinmathbb{R}setminus mathbb {Q} $. $f$ is continuous and differentiable at $0$ and at other points it is neither continuous nor differentiable.
– Paramanand Singh
Nov 18 at 14:47