Mixing
How do you prove the following about a function approaching infinity at some point?
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How do you prove that the derivative of a continuous function which approaches infinity from the right or the left of some point $x_0$ also approaches infinity? I suppose the definition of the limit needs to be applied somehow, but I don't know how. Thank you.
$lim_{xrightarrow x_0^+}{f(x)} = infty rightarrow lim_{xrightarrow x_0^+}{f'(x)} = -infty$
Is $$f'(x)=-dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$$ a counterexample or not?
calculus limits
edited Jan 6 at 22:34
RRL
50.2k42573
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
$endgroup$
add a comment |
$begingroup$
How do you prove that the derivative of a continuous function which approaches infinity from the right or the left of some point $x_0$ also approaches infinity? I suppose the definition of the limit needs to be applied somehow, but I don't know how. Thank you.
$lim_{xrightarrow x_0^+}{f(x)} = infty rightarrow lim_{xrightarrow x_0^+}{f'(x)} = -infty$
Is $$f'(x)=-dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$$ a counterexample or not?
calculus limits
edited Jan 6 at 22:34
RRL
50.2k42573
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
$endgroup$
$begingroup$
This is false. You cannot prove it.
$endgroup$
– Crostul
Jan 6 at 19:23
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Could you clarify why?
$endgroup$
– Andrew Blitz
Jan 6 at 20:03
1
$begingroup$
@AndrewBlitz: Assuming there is a derivative it is true that $liminf_{x to x_0+} f'(x) = -infty$ but I believe it is possible that the limit may not exist.
$endgroup$
– RRL
Jan 6 at 20:12
$begingroup$
What if $f'(x) = dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$?
$endgroup$
– Andrew Blitz
Jan 6 at 22:22
add a comment |
$begingroup$
How do you prove that the derivative of a continuous function which approaches infinity from the right or the left of some point $x_0$ also approaches infinity? I suppose the definition of the limit needs to be applied somehow, but I don't know how. Thank you.
$lim_{xrightarrow x_0^+}{f(x)} = infty rightarrow lim_{xrightarrow x_0^+}{f'(x)} = -infty$
Is $$f'(x)=-dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$$ a counterexample or not?
calculus limits
edited Jan 6 at 22:34
RRL
50.2k42573
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
$endgroup$
How do you prove that the derivative of a continuous function which approaches infinity from the right or the left of some point $x_0$ also approaches infinity? I suppose the definition of the limit needs to be applied somehow, but I don't know how. Thank you.
$lim_{xrightarrow x_0^+}{f(x)} = infty rightarrow lim_{xrightarrow x_0^+}{f'(x)} = -infty$
Is $$f'(x)=-dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$$ a counterexample or not?
calculus limits
calculus limits
edited Jan 6 at 22:34
RRL
50.2k42573
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
edited Jan 6 at 22:34
RRL
50.2k42573
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
edited Jan 6 at 22:34
RRL
50.2k42573
edited Jan 6 at 22:34
RRL
50.2k42573
edited Jan 6 at 22:34
RRL
50.2k42573
50.2k42573
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
asked Jan 6 at 19:19
Andrew BlitzAndrew Blitz
414
414
$begingroup$
This is false. You cannot prove it.
$endgroup$
– Crostul
Jan 6 at 19:23
$begingroup$
Could you clarify why?
$endgroup$
– Andrew Blitz
Jan 6 at 20:03
1
$begingroup$
@AndrewBlitz: Assuming there is a derivative it is true that $liminf_{x to x_0+} f'(x) = -infty$ but I believe it is possible that the limit may not exist.
$endgroup$
– RRL
Jan 6 at 20:12
$begingroup$
What if $f'(x) = dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$?
$endgroup$
– Andrew Blitz
Jan 6 at 22:22
add a comment |
$begingroup$
This is false. You cannot prove it.
$endgroup$
– Crostul
Jan 6 at 19:23
$begingroup$
Could you clarify why?
$endgroup$
– Andrew Blitz
Jan 6 at 20:03
1
$begingroup$
@AndrewBlitz: Assuming there is a derivative it is true that $liminf_{x to x_0+} f'(x) = -infty$ but I believe it is possible that the limit may not exist.
$endgroup$
– RRL
Jan 6 at 20:12
$begingroup$
What if $f'(x) = dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$?
$endgroup$
– Andrew Blitz
Jan 6 at 22:22
$begingroup$
This is false. You cannot prove it.
$endgroup$
– Crostul
Jan 6 at 19:23
$begingroup$
This is false. You cannot prove it.
$endgroup$
– Crostul
Jan 6 at 19:23
$begingroup$
Could you clarify why?
$endgroup$
– Andrew Blitz
Jan 6 at 20:03
$begingroup$
Could you clarify why?
$endgroup$
– Andrew Blitz
Jan 6 at 20:03
1
1
$begingroup$
@AndrewBlitz: Assuming there is a derivative it is true that $liminf_{x to x_0+} f'(x) = -infty$ but I believe it is possible that the limit may not exist.
$endgroup$
– RRL
Jan 6 at 20:12
$begingroup$
@AndrewBlitz: Assuming there is a derivative it is true that $liminf_{x to x_0+} f'(x) = -infty$ but I believe it is possible that the limit may not exist.
$endgroup$
– RRL
Jan 6 at 20:12
$begingroup$
What if $f'(x) = dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$?
$endgroup$
– Andrew Blitz
Jan 6 at 22:22
$begingroup$
What if $f'(x) = dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$?
$endgroup$
– Andrew Blitz
Jan 6 at 22:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I doubt that the statement is true. For example,
$$
f(x) = int_x^1 frac{1+cos(frac{pi}{t})}{t}dt
$$ has vanishing derivative at $x=frac{1}{(2k+1)}$, $kge 1$ but for all $epsilon <frac{1}{N}$, we have
$$
f(epsilon)geint_{frac{1}{N}}^1 frac{1+cos(frac{pi}{t})}{t}dt = int_{1}^N frac{1+cos(pi u)}{u}duge sum_{n=1}^{N-1} frac{1}{n+1}int_{n}^{n+1}[1+cos(pi u)] du =sum_{n=1}^{N-1}frac{1}{n+1},
$$which is saying that $lim_{xto 0^+} f(x) =infty.$
answered Jan 6 at 21:55
SongSong
10.4k627
$endgroup$
add a comment |
$begingroup$
It is true that $liminf_{x to x_0+} f'(x) = - infty$ but it is not necessary that $lim_{x to x_0} f'(x)$ exists.
To prove the first statement take a sequence $x_n to x_0+$ where $f(x_n) to +infty$. Take a fixed $y$ such that $x_0 < x_n < y$ and $f(x_n) > f(y)$ for all $n$. By the MVT there is a sequence $xi_n$ such that
$$f'(xi_n) = frac{f(y) - f(x_n)}{y - x_n} < frac{f(y) - f(x_n)}{y - x_0}$$
and the RHS converges to $-infty$ as $n to infty$.
For a counterexample to existence of the limit, take $x_0 = 0$ and $f(x) = - log(x^2 sin frac{1}{x})$. Here we have $f(x) to +infty $ as $x to 0+$, and
$$f'(x) = frac{2}{x} - frac{cot frac{1}{x}}{x^2} = frac{1}{x^2}left(2x - cot frac{1}{x} right)$$
Although $f'$ is unbounded in a neighborhood of $0$, it oscillates perpetually (passing through $0$) and the limit does not exist. Note that $cot y - 2/y$ has infinitely many zeros in $(1, infty)$.
answered Jan 6 at 22:02
RRLRRL
50.2k42573
$endgroup$
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
add a comment |
$begingroup$
You need f to be differentiable and for the limit to exist. For $x_0<x<x+1$ select $x_0<y<x$ with $f(y)>f(x)+n$. Apply the mvt and let $ntoinfty$. Then we get $liminf_{xto x_0^+}f'(x)=-infty$, which since the limit exists means $lim_{xto x_0^+}f'(x)=-infty$.
answered Jan 6 at 19:30
Ben WBen W
2,234615
$endgroup$
1
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
$endgroup$
– RRL
Jan 6 at 20:10
$begingroup$
@RRL oops! Fixed.
$endgroup$
– Ben W
Jan 7 at 1:02
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I doubt that the statement is true. For example,
$$
f(x) = int_x^1 frac{1+cos(frac{pi}{t})}{t}dt
$$ has vanishing derivative at $x=frac{1}{(2k+1)}$, $kge 1$ but for all $epsilon <frac{1}{N}$, we have
$$
f(epsilon)geint_{frac{1}{N}}^1 frac{1+cos(frac{pi}{t})}{t}dt = int_{1}^N frac{1+cos(pi u)}{u}duge sum_{n=1}^{N-1} frac{1}{n+1}int_{n}^{n+1}[1+cos(pi u)] du =sum_{n=1}^{N-1}frac{1}{n+1},
$$which is saying that $lim_{xto 0^+} f(x) =infty.$
answered Jan 6 at 21:55
SongSong
10.4k627
$endgroup$
add a comment |
$begingroup$
I doubt that the statement is true. For example,
$$
f(x) = int_x^1 frac{1+cos(frac{pi}{t})}{t}dt
$$ has vanishing derivative at $x=frac{1}{(2k+1)}$, $kge 1$ but for all $epsilon <frac{1}{N}$, we have
$$
f(epsilon)geint_{frac{1}{N}}^1 frac{1+cos(frac{pi}{t})}{t}dt = int_{1}^N frac{1+cos(pi u)}{u}duge sum_{n=1}^{N-1} frac{1}{n+1}int_{n}^{n+1}[1+cos(pi u)] du =sum_{n=1}^{N-1}frac{1}{n+1},
$$which is saying that $lim_{xto 0^+} f(x) =infty.$
answered Jan 6 at 21:55
SongSong
10.4k627
$endgroup$
add a comment |
$begingroup$
I doubt that the statement is true. For example,
$$
f(x) = int_x^1 frac{1+cos(frac{pi}{t})}{t}dt
$$ has vanishing derivative at $x=frac{1}{(2k+1)}$, $kge 1$ but for all $epsilon <frac{1}{N}$, we have
$$
f(epsilon)geint_{frac{1}{N}}^1 frac{1+cos(frac{pi}{t})}{t}dt = int_{1}^N frac{1+cos(pi u)}{u}duge sum_{n=1}^{N-1} frac{1}{n+1}int_{n}^{n+1}[1+cos(pi u)] du =sum_{n=1}^{N-1}frac{1}{n+1},
$$which is saying that $lim_{xto 0^+} f(x) =infty.$
answered Jan 6 at 21:55
SongSong
10.4k627
$endgroup$
I doubt that the statement is true. For example,
$$
f(x) = int_x^1 frac{1+cos(frac{pi}{t})}{t}dt
$$ has vanishing derivative at $x=frac{1}{(2k+1)}$, $kge 1$ but for all $epsilon <frac{1}{N}$, we have
$$
f(epsilon)geint_{frac{1}{N}}^1 frac{1+cos(frac{pi}{t})}{t}dt = int_{1}^N frac{1+cos(pi u)}{u}duge sum_{n=1}^{N-1} frac{1}{n+1}int_{n}^{n+1}[1+cos(pi u)] du =sum_{n=1}^{N-1}frac{1}{n+1},
$$which is saying that $lim_{xto 0^+} f(x) =infty.$
answered Jan 6 at 21:55
SongSong
10.4k627
answered Jan 6 at 21:55
SongSong
10.4k627
answered Jan 6 at 21:55
SongSong
10.4k627
answered Jan 6 at 21:55
SongSong
10.4k627
10.4k627
add a comment |
add a comment |
$begingroup$
It is true that $liminf_{x to x_0+} f'(x) = - infty$ but it is not necessary that $lim_{x to x_0} f'(x)$ exists.
To prove the first statement take a sequence $x_n to x_0+$ where $f(x_n) to +infty$. Take a fixed $y$ such that $x_0 < x_n < y$ and $f(x_n) > f(y)$ for all $n$. By the MVT there is a sequence $xi_n$ such that
$$f'(xi_n) = frac{f(y) - f(x_n)}{y - x_n} < frac{f(y) - f(x_n)}{y - x_0}$$
and the RHS converges to $-infty$ as $n to infty$.
For a counterexample to existence of the limit, take $x_0 = 0$ and $f(x) = - log(x^2 sin frac{1}{x})$. Here we have $f(x) to +infty $ as $x to 0+$, and
$$f'(x) = frac{2}{x} - frac{cot frac{1}{x}}{x^2} = frac{1}{x^2}left(2x - cot frac{1}{x} right)$$
Although $f'$ is unbounded in a neighborhood of $0$, it oscillates perpetually (passing through $0$) and the limit does not exist. Note that $cot y - 2/y$ has infinitely many zeros in $(1, infty)$.
answered Jan 6 at 22:02
RRLRRL
50.2k42573
$endgroup$
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
add a comment |
$begingroup$
It is true that $liminf_{x to x_0+} f'(x) = - infty$ but it is not necessary that $lim_{x to x_0} f'(x)$ exists.
To prove the first statement take a sequence $x_n to x_0+$ where $f(x_n) to +infty$. Take a fixed $y$ such that $x_0 < x_n < y$ and $f(x_n) > f(y)$ for all $n$. By the MVT there is a sequence $xi_n$ such that
$$f'(xi_n) = frac{f(y) - f(x_n)}{y - x_n} < frac{f(y) - f(x_n)}{y - x_0}$$
and the RHS converges to $-infty$ as $n to infty$.
For a counterexample to existence of the limit, take $x_0 = 0$ and $f(x) = - log(x^2 sin frac{1}{x})$. Here we have $f(x) to +infty $ as $x to 0+$, and
$$f'(x) = frac{2}{x} - frac{cot frac{1}{x}}{x^2} = frac{1}{x^2}left(2x - cot frac{1}{x} right)$$
Although $f'$ is unbounded in a neighborhood of $0$, it oscillates perpetually (passing through $0$) and the limit does not exist. Note that $cot y - 2/y$ has infinitely many zeros in $(1, infty)$.
answered Jan 6 at 22:02
RRLRRL
50.2k42573
$endgroup$
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
add a comment |
$begingroup$
It is true that $liminf_{x to x_0+} f'(x) = - infty$ but it is not necessary that $lim_{x to x_0} f'(x)$ exists.
To prove the first statement take a sequence $x_n to x_0+$ where $f(x_n) to +infty$. Take a fixed $y$ such that $x_0 < x_n < y$ and $f(x_n) > f(y)$ for all $n$. By the MVT there is a sequence $xi_n$ such that
$$f'(xi_n) = frac{f(y) - f(x_n)}{y - x_n} < frac{f(y) - f(x_n)}{y - x_0}$$
and the RHS converges to $-infty$ as $n to infty$.
For a counterexample to existence of the limit, take $x_0 = 0$ and $f(x) = - log(x^2 sin frac{1}{x})$. Here we have $f(x) to +infty $ as $x to 0+$, and
$$f'(x) = frac{2}{x} - frac{cot frac{1}{x}}{x^2} = frac{1}{x^2}left(2x - cot frac{1}{x} right)$$
Although $f'$ is unbounded in a neighborhood of $0$, it oscillates perpetually (passing through $0$) and the limit does not exist. Note that $cot y - 2/y$ has infinitely many zeros in $(1, infty)$.
answered Jan 6 at 22:02
RRLRRL
50.2k42573
$endgroup$
It is true that $liminf_{x to x_0+} f'(x) = - infty$ but it is not necessary that $lim_{x to x_0} f'(x)$ exists.
To prove the first statement take a sequence $x_n to x_0+$ where $f(x_n) to +infty$. Take a fixed $y$ such that $x_0 < x_n < y$ and $f(x_n) > f(y)$ for all $n$. By the MVT there is a sequence $xi_n$ such that
$$f'(xi_n) = frac{f(y) - f(x_n)}{y - x_n} < frac{f(y) - f(x_n)}{y - x_0}$$
and the RHS converges to $-infty$ as $n to infty$.
For a counterexample to existence of the limit, take $x_0 = 0$ and $f(x) = - log(x^2 sin frac{1}{x})$. Here we have $f(x) to +infty $ as $x to 0+$, and
$$f'(x) = frac{2}{x} - frac{cot frac{1}{x}}{x^2} = frac{1}{x^2}left(2x - cot frac{1}{x} right)$$
Although $f'$ is unbounded in a neighborhood of $0$, it oscillates perpetually (passing through $0$) and the limit does not exist. Note that $cot y - 2/y$ has infinitely many zeros in $(1, infty)$.
answered Jan 6 at 22:02
RRLRRL
50.2k42573
edited Jan 6 at 22:27
answered Jan 6 at 22:02
RRLRRL
50.2k42573
answered Jan 6 at 22:02
RRLRRL
50.2k42573
answered Jan 6 at 22:02
RRLRRL
50.2k42573
50.2k42573
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
add a comment |
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
$begingroup$
Good catch! Sorry I missed it.
$endgroup$
– Ben W
Jan 7 at 1:01
add a comment |
$begingroup$
You need f to be differentiable and for the limit to exist. For $x_0<x<x+1$ select $x_0<y<x$ with $f(y)>f(x)+n$. Apply the mvt and let $ntoinfty$. Then we get $liminf_{xto x_0^+}f'(x)=-infty$, which since the limit exists means $lim_{xto x_0^+}f'(x)=-infty$.
answered Jan 6 at 19:30
Ben WBen W
2,234615
$endgroup$
1
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
$endgroup$
– RRL
Jan 6 at 20:10
$begingroup$
@RRL oops! Fixed.
$endgroup$
– Ben W
Jan 7 at 1:02
add a comment |
$begingroup$
You need f to be differentiable and for the limit to exist. For $x_0<x<x+1$ select $x_0<y<x$ with $f(y)>f(x)+n$. Apply the mvt and let $ntoinfty$. Then we get $liminf_{xto x_0^+}f'(x)=-infty$, which since the limit exists means $lim_{xto x_0^+}f'(x)=-infty$.
answered Jan 6 at 19:30
Ben WBen W
2,234615
$endgroup$
1
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
$endgroup$
– RRL
Jan 6 at 20:10
$begingroup$
@RRL oops! Fixed.
$endgroup$
– Ben W
Jan 7 at 1:02
add a comment |
$begingroup$
You need f to be differentiable and for the limit to exist. For $x_0<x<x+1$ select $x_0<y<x$ with $f(y)>f(x)+n$. Apply the mvt and let $ntoinfty$. Then we get $liminf_{xto x_0^+}f'(x)=-infty$, which since the limit exists means $lim_{xto x_0^+}f'(x)=-infty$.
answered Jan 6 at 19:30
Ben WBen W
2,234615
$endgroup$
You need f to be differentiable and for the limit to exist. For $x_0<x<x+1$ select $x_0<y<x$ with $f(y)>f(x)+n$. Apply the mvt and let $ntoinfty$. Then we get $liminf_{xto x_0^+}f'(x)=-infty$, which since the limit exists means $lim_{xto x_0^+}f'(x)=-infty$.
answered Jan 6 at 19:30
Ben WBen W
2,234615
edited Jan 7 at 1:02
answered Jan 6 at 19:30
Ben WBen W
2,234615
answered Jan 6 at 19:30
Ben WBen W
2,234615
answered Jan 6 at 19:30
Ben WBen W
2,234615
2,234615
1
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
$endgroup$
– RRL
Jan 6 at 20:10
$begingroup$
@RRL oops! Fixed.
$endgroup$
– Ben W
Jan 7 at 1:02
add a comment |
1
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
$endgroup$
– RRL
Jan 6 at 20:10
$begingroup$
@RRL oops! Fixed.
$endgroup$
– Ben W
Jan 7 at 1:02
1
1
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
$endgroup$
– RRL
Jan 6 at 20:10
$begingroup$
Could produce a few more steps to show that this proves $lim_{x to x_0+}f'(x) = -infty$ and not just the existence of a sequence $xi_n to x_0+$ where $f'(xi_n) to -infty$ (i.e., unbounded but with non-existent limit)
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– RRL
Jan 6 at 20:10
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@RRL oops! Fixed.
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– Ben W
Jan 7 at 1:02
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@RRL oops! Fixed.
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– Ben W
Jan 7 at 1:02
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$begingroup$
This is false. You cannot prove it.
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– Crostul
Jan 6 at 19:23
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Could you clarify why?
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– Andrew Blitz
Jan 6 at 20:03
1
$begingroup$
@AndrewBlitz: Assuming there is a derivative it is true that $liminf_{x to x_0+} f'(x) = -infty$ but I believe it is possible that the limit may not exist.
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– RRL
Jan 6 at 20:12
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What if $f'(x) = dfrac{1}{x-x_0}(sin(dfrac{1}{x-x_0})+1)$?
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– Andrew Blitz
Jan 6 at 22:22