Derivative of continuous function exists if limit of derivative exists
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I'm stuck on this old qualifier problem. I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)
Let $f:mathbb{R} mapsto mathbb{R}$ be continuous. Suppose $f$ is differentiable away from $0$ and lim$_{x to 0} f^prime(x)$ exists. Show $f^prime(0)$ exists.
calculus real-analysis analysis
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
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add a comment |
$begingroup$
I'm stuck on this old qualifier problem. I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)
Let $f:mathbb{R} mapsto mathbb{R}$ be continuous. Suppose $f$ is differentiable away from $0$ and lim$_{x to 0} f^prime(x)$ exists. Show $f^prime(0)$ exists.
calculus real-analysis analysis
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
$endgroup$
add a comment |
$begingroup$
I'm stuck on this old qualifier problem. I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)
Let $f:mathbb{R} mapsto mathbb{R}$ be continuous. Suppose $f$ is differentiable away from $0$ and lim$_{x to 0} f^prime(x)$ exists. Show $f^prime(0)$ exists.
calculus real-analysis analysis
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
$endgroup$
I'm stuck on this old qualifier problem. I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)
Let $f:mathbb{R} mapsto mathbb{R}$ be continuous. Suppose $f$ is differentiable away from $0$ and lim$_{x to 0} f^prime(x)$ exists. Show $f^prime(0)$ exists.
calculus real-analysis analysis
calculus real-analysis analysis
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
asked Jul 5 '14 at 14:46
StrangerLoopStrangerLoop
74649
74649
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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By the mean value theorem, there is a $c_xin (0,x)$ resp. $c_xin (x,0)$, depending on whether $x > 0$ or $x < 0$, such that
$$frac{f(x)-f(0)}{x} = f'(c_x).$$
As $xto 0$, by the squeeze lemma, also $c_xto 0$, hence
$$lim_{xto 0} frac{f(x)-f(0)}{x} = lim_{xto 0}f'(c_x)$$
exists.
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
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$begingroup$
Nice answer! Thanks.
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– StrangerLoop
Jul 5 '14 at 14:53
add a comment |
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Not a proof, but an example I use with my students...
Consider the derivative of an absolute value function.
f(x)=abs(x)
The derivative is f(x)={1, x>0
{-1, x<0
Therefore, the function is continuous and differentiable, but the derivative is not continuous.
answered Mar 28 '18 at 12:35
user546542user546542
1
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$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By the mean value theorem, there is a $c_xin (0,x)$ resp. $c_xin (x,0)$, depending on whether $x > 0$ or $x < 0$, such that
$$frac{f(x)-f(0)}{x} = f'(c_x).$$
As $xto 0$, by the squeeze lemma, also $c_xto 0$, hence
$$lim_{xto 0} frac{f(x)-f(0)}{x} = lim_{xto 0}f'(c_x)$$
exists.
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
$endgroup$
$begingroup$
Nice answer! Thanks.
$endgroup$
– StrangerLoop
Jul 5 '14 at 14:53
add a comment |
$begingroup$
By the mean value theorem, there is a $c_xin (0,x)$ resp. $c_xin (x,0)$, depending on whether $x > 0$ or $x < 0$, such that
$$frac{f(x)-f(0)}{x} = f'(c_x).$$
As $xto 0$, by the squeeze lemma, also $c_xto 0$, hence
$$lim_{xto 0} frac{f(x)-f(0)}{x} = lim_{xto 0}f'(c_x)$$
exists.
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
$endgroup$
$begingroup$
Nice answer! Thanks.
$endgroup$
– StrangerLoop
Jul 5 '14 at 14:53
add a comment |
$begingroup$
By the mean value theorem, there is a $c_xin (0,x)$ resp. $c_xin (x,0)$, depending on whether $x > 0$ or $x < 0$, such that
$$frac{f(x)-f(0)}{x} = f'(c_x).$$
As $xto 0$, by the squeeze lemma, also $c_xto 0$, hence
$$lim_{xto 0} frac{f(x)-f(0)}{x} = lim_{xto 0}f'(c_x)$$
exists.
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
$endgroup$
By the mean value theorem, there is a $c_xin (0,x)$ resp. $c_xin (x,0)$, depending on whether $x > 0$ or $x < 0$, such that
$$frac{f(x)-f(0)}{x} = f'(c_x).$$
As $xto 0$, by the squeeze lemma, also $c_xto 0$, hence
$$lim_{xto 0} frac{f(x)-f(0)}{x} = lim_{xto 0}f'(c_x)$$
exists.
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
answered Jul 5 '14 at 14:50
Daniel Fischer♦Daniel Fischer
173k16165285
173k16165285
$begingroup$
Nice answer! Thanks.
$endgroup$
– StrangerLoop
Jul 5 '14 at 14:53
add a comment |
$begingroup$
Nice answer! Thanks.
$endgroup$
– StrangerLoop
Jul 5 '14 at 14:53
$begingroup$
Nice answer! Thanks.
$endgroup$
– StrangerLoop
Jul 5 '14 at 14:53
$begingroup$
Nice answer! Thanks.
$endgroup$
– StrangerLoop
Jul 5 '14 at 14:53
add a comment |
$begingroup$
Not a proof, but an example I use with my students...
Consider the derivative of an absolute value function.
f(x)=abs(x)
The derivative is f(x)={1, x>0
{-1, x<0
Therefore, the function is continuous and differentiable, but the derivative is not continuous.
answered Mar 28 '18 at 12:35
user546542user546542
1
$endgroup$
$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
add a comment |
$begingroup$
Not a proof, but an example I use with my students...
Consider the derivative of an absolute value function.
f(x)=abs(x)
The derivative is f(x)={1, x>0
{-1, x<0
Therefore, the function is continuous and differentiable, but the derivative is not continuous.
answered Mar 28 '18 at 12:35
user546542user546542
1
$endgroup$
$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
add a comment |
$begingroup$
Not a proof, but an example I use with my students...
Consider the derivative of an absolute value function.
f(x)=abs(x)
The derivative is f(x)={1, x>0
{-1, x<0
Therefore, the function is continuous and differentiable, but the derivative is not continuous.
answered Mar 28 '18 at 12:35
user546542user546542
1
$endgroup$
Not a proof, but an example I use with my students...
Consider the derivative of an absolute value function.
f(x)=abs(x)
The derivative is f(x)={1, x>0
{-1, x<0
Therefore, the function is continuous and differentiable, but the derivative is not continuous.
answered Mar 28 '18 at 12:35
user546542user546542
1
answered Mar 28 '18 at 12:35
user546542user546542
1
answered Mar 28 '18 at 12:35
user546542user546542
1
answered Mar 28 '18 at 12:35
user546542user546542
1
1
$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
add a comment |
$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
$begingroup$
The question assumes that the limit of the derivative exists. This is not the case here.
$endgroup$
– Arnaud D.
Mar 28 '18 at 14:30
add a comment |
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