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$int _0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f '(x)$ exists. Counter...
Can anyone give me a counter example of the statement
If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.
My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.
1) $1/x^2$
2) $-1/x^2$
3) $x= 1$.
Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.
I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.
limits analysis derivatives proof-verification continuity
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
asked Nov 22 '18 at 4:57
cmicmi
1,015212
add a comment |
Can anyone give me a counter example of the statement
If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.
My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.
1) $1/x^2$
2) $-1/x^2$
3) $x= 1$.
Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.
I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.
limits analysis derivatives proof-verification continuity
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
asked Nov 22 '18 at 4:57
cmicmi
1,015212
add a comment |
Can anyone give me a counter example of the statement
If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.
My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.
1) $1/x^2$
2) $-1/x^2$
3) $x= 1$.
Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.
I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.
limits analysis derivatives proof-verification continuity
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
asked Nov 22 '18 at 4:57
cmicmi
1,015212
Can anyone give me a counter example of the statement
If $int_0^infty f(x) $ exists and $f(x)$ is differentiable then $lim _{x to infty} f'(x)$ exists.
My attempt: I have thought one. First I draw $1/x^2$ in the first quadrant and $-1/x^2$ in the fourth quadrant. The area under the following curves are finite.
1) $1/x^2$
2) $-1/x^2$
3) $x= 1$.
Now I have drawn infinite number of $y = x+c $ at equal distances in that region. Then I joined those infinite lines by some smooth curve so that the curve remains differentiable. Now I think this function can be a counter example.
I am uploading one picture of my attempt. Can anyone please check it and if possible suggest me a better function.
limits analysis derivatives proof-verification continuity
limits analysis derivatives proof-verification continuity
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
asked Nov 22 '18 at 4:57
cmicmi
1,015212
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
asked Nov 22 '18 at 4:57
cmicmi
1,015212
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
edited Nov 22 '18 at 5:07
gt6989b
33.3k22452
33.3k22452
asked Nov 22 '18 at 4:57
cmicmi
1,015212
asked Nov 22 '18 at 4:57
cmicmi
1,015212
asked Nov 22 '18 at 4:57
cmicmi
1,015212
1,015212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Yep, this works perfectly! You can rigorize this sort of idea by defining some function like
$$frac{sinleft(x^{10}right)}{x^2}$$
(where the exponent of $10$ is simply to make sure our function oscillates fast enough).
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
1
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
1
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yep, this works perfectly! You can rigorize this sort of idea by defining some function like
$$frac{sinleft(x^{10}right)}{x^2}$$
(where the exponent of $10$ is simply to make sure our function oscillates fast enough).
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
1
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
1
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
add a comment |
Yep, this works perfectly! You can rigorize this sort of idea by defining some function like
$$frac{sinleft(x^{10}right)}{x^2}$$
(where the exponent of $10$ is simply to make sure our function oscillates fast enough).
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
1
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
1
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
add a comment |
Yep, this works perfectly! You can rigorize this sort of idea by defining some function like
$$frac{sinleft(x^{10}right)}{x^2}$$
(where the exponent of $10$ is simply to make sure our function oscillates fast enough).
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
Yep, this works perfectly! You can rigorize this sort of idea by defining some function like
$$frac{sinleft(x^{10}right)}{x^2}$$
(where the exponent of $10$ is simply to make sure our function oscillates fast enough).
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
answered Nov 22 '18 at 5:02
Carl SchildkrautCarl Schildkraut
11.2k11441
11.2k11441
1
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
1
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
add a comment |
1
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
1
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
1
1
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
Actually simply $sin (x^2)$ suffices.
– Szeto
Nov 22 '18 at 5:24
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
@Szeto I imagined lower exponents sufficed, but I didn't want to do out the calculations :-)
– Carl Schildkraut
Nov 22 '18 at 5:47
1
1
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
Well, this function is in my memory because of the Fresnel’s integral, which is the first integral I see that converges while its integrand does not vanish. :)
– Szeto
Nov 22 '18 at 5:49
add a comment |
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