Can we apply 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 it's derivative may not be continuous at some point. One of many counter examples 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
asked 1 hour ago
DhvanitDhvanit
1088
$endgroup$
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 it's derivative may not be continuous at some point. One of many counter examples 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
asked 1 hour ago
DhvanitDhvanit
1088
$endgroup$
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 it's derivative may not be continuous at some point. One of many counter examples 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
asked 1 hour ago
DhvanitDhvanit
1088
$endgroup$
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 it's derivative may not be continuous at some point. One of many counter examples 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
limits derivatives
asked 1 hour ago
DhvanitDhvanit
1088
asked 1 hour ago
DhvanitDhvanit
1088
asked 1 hour ago
DhvanitDhvanit
1088
asked 1 hour ago
DhvanitDhvanit
1088
asked 1 hour ago
DhvanitDhvanit
1088
1088
add a comment |
add a comment |
2 Answers
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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 1 hour ago
HelmutHelmut
739128
$endgroup$
$begingroup$
@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
1 hour 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
52 mins 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 1 hour ago
mihaildmihaild
5689
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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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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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 1 hour ago
HelmutHelmut
739128
$endgroup$
$begingroup$
@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
1 hour 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
52 mins 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 1 hour ago
HelmutHelmut
739128
$endgroup$
$begingroup$
@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
1 hour 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
52 mins 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 1 hour ago
HelmutHelmut
739128
$endgroup$
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 1 hour ago
HelmutHelmut
739128
answered 1 hour ago
HelmutHelmut
739128
answered 1 hour ago
HelmutHelmut
739128
answered 1 hour ago
HelmutHelmut
739128
739128
$begingroup$
@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
1 hour 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
52 mins 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
1 hour 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
52 mins 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
1 hour 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
1 hour 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
52 mins 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
52 mins 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 1 hour ago
mihaildmihaild
5689
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 1 hour ago
mihaildmihaild
5689
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 1 hour ago
mihaildmihaild
5689
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 1 hour ago
mihaildmihaild
5689
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 1 hour ago
mihaildmihaild
5689
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 1 hour ago
mihaildmihaild
5689
answered 1 hour ago
mihaildmihaild
5689
5689
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 |
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