Can we apply L'Hospital's rule where the derivative is not continuous? The 2019 Stack Overflow...
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Can we apply L'Hospital's rule where the derivative is not continuous?
The 2019 Stack Overflow Developer Survey Results Are InResilient L'Hospital's rule questionIs this a valid use of l'Hospital's Rule? Can it be used recursively?Simple Derivation of Functional Equation Question (L'Hospital's Rule)L'Hôpital's rule and Difference QuotientsL'Hôpital's rule does not apply?!Fake proof for “differentiability implies continuous derivative”: reviewL'Hospital's rule helpCan a function be differentiable on an interval, but not continuously differentiable somewhere besides an oscillating discontinuity in the derivative?Compute the limit without L'Hospital's ruleProof of L'Hospital's Rule
$begingroup$
My doubt arises due to the following :
We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h}$$
Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $
So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h} = lim_{h rightarrow 0} frac {f'(a+h)}{1}$$
Which implies that
$$f'(a) = lim_{h rightarrow 0} f'(a+h)$$
Which means that the function $f'(x)$ is continuous at $x=a$
But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
$$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
Whose derivative isn't continuous at $0$
So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?
If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?
limits derivatives continuity
edited 1 hour ago
200_success
668515
asked 9 hours ago
DhvanitDhvanit
1239
$endgroup$
add a comment |
$begingroup$
My doubt arises due to the following :
We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h}$$
Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $
So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h} = lim_{h rightarrow 0} frac {f'(a+h)}{1}$$
Which implies that
$$f'(a) = lim_{h rightarrow 0} f'(a+h)$$
Which means that the function $f'(x)$ is continuous at $x=a$
But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
$$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
Whose derivative isn't continuous at $0$
So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?
If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?
limits derivatives continuity
edited 1 hour ago
200_success
668515
asked 9 hours ago
DhvanitDhvanit
1239
$endgroup$
add a comment |
$begingroup$
My doubt arises due to the following :
We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h}$$
Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $
So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h} = lim_{h rightarrow 0} frac {f'(a+h)}{1}$$
Which implies that
$$f'(a) = lim_{h rightarrow 0} f'(a+h)$$
Which means that the function $f'(x)$ is continuous at $x=a$
But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
$$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
Whose derivative isn't continuous at $0$
So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?
If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?
limits derivatives continuity
edited 1 hour ago
200_success
668515
asked 9 hours ago
DhvanitDhvanit
1239
$endgroup$
My doubt arises due to the following :
We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h}$$
Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $
So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
$$f'(a) = lim_{h rightarrow 0} frac {f(a+h) - f(a)}{h} = lim_{h rightarrow 0} frac {f'(a+h)}{1}$$
Which implies that
$$f'(a) = lim_{h rightarrow 0} f'(a+h)$$
Which means that the function $f'(x)$ is continuous at $x=a$
But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
$$f(x) = begin{cases} 0 text{ ; if x=0} \ x^2 sin frac{1}{x} text{; if x $neq$ 0 } end{cases}$$
Whose derivative isn't continuous at $0$
So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?
If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?
limits derivatives continuity
limits derivatives continuity
edited 1 hour ago
200_success
668515
asked 9 hours ago
DhvanitDhvanit
1239
edited 1 hour ago
200_success
668515
asked 9 hours ago
DhvanitDhvanit
1239
edited 1 hour ago
200_success
668515
edited 1 hour ago
200_success
668515
edited 1 hour ago
200_success
668515
668515
asked 9 hours ago
DhvanitDhvanit
1239
asked 9 hours ago
DhvanitDhvanit
1239
asked 9 hours ago
DhvanitDhvanit
1239
1239
add a comment |
add a comment |
3 Answers
3
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$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. It does not say anything about the existence of the former limit.
answered 9 hours ago
HelmutHelmut
789128
$endgroup$
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.
answered 9 hours ago
mihaildmihaild
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
The derivative value exists if:
1. the left-side (-0) derivative exists
2. the right-side (+0) derivative exists
3. and they are the same/identical .
In your case they are not identical.
answered 7 hours ago
user9user9
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. It does not say anything about the existence of the former limit.
answered 9 hours ago
HelmutHelmut
789128
$endgroup$
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
add a comment |
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. It does not say anything about the existence of the former limit.
answered 9 hours ago
HelmutHelmut
789128
$endgroup$
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
add a comment |
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. It does not say anything about the existence of the former limit.
answered 9 hours ago
HelmutHelmut
789128
$endgroup$
L'Hospital's rule says under certain conditions: IF $lim_{hto 0} frac{f'(h)}{g'(h)}=c$ exists, then also $lim_{hto 0} frac{f(h)}{g(h)}=c$. It does not say anything about the existence of the former limit.
answered 9 hours ago
HelmutHelmut
789128
answered 9 hours ago
HelmutHelmut
789128
answered 9 hours ago
HelmutHelmut
789128
answered 9 hours ago
HelmutHelmut
789128
789128
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
add a comment |
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
@Dhvanit when $lim_{hto0}frac{f'(h)}{g'(h)}$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
9 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_{hto 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition).
$endgroup$
– Ingix
8 hours ago
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.
answered 9 hours ago
mihaildmihaild
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.
answered 9 hours ago
mihaildmihaild
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.
answered 9 hours ago
mihaildmihaild
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim frac{f'(x)}{g'(x)}$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.
answered 9 hours ago
mihaildmihaild
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 9 hours ago
mihaildmihaild
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 9 hours ago
mihaildmihaild
56810
answered 9 hours ago
mihaildmihaild
56810
56810
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mihaild is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
The derivative value exists if:
1. the left-side (-0) derivative exists
2. the right-side (+0) derivative exists
3. and they are the same/identical .
In your case they are not identical.
answered 7 hours ago
user9user9
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The derivative value exists if:
1. the left-side (-0) derivative exists
2. the right-side (+0) derivative exists
3. and they are the same/identical .
In your case they are not identical.
answered 7 hours ago
user9user9
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The derivative value exists if:
1. the left-side (-0) derivative exists
2. the right-side (+0) derivative exists
3. and they are the same/identical .
In your case they are not identical.
answered 7 hours ago
user9user9
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The derivative value exists if:
1. the left-side (-0) derivative exists
2. the right-side (+0) derivative exists
3. and they are the same/identical .
In your case they are not identical.
answered 7 hours ago
user9user9
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
user9user9
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
user9user9
1
answered 7 hours ago
user9user9
1
1
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user9 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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