Can we apply L'Hospital's rule? The 2019 Stack Overflow Developer Survey Results Are InResilient L'Hospital's rule questionUsing L'Hospital's Rule to evaluate limit to infinityIs 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 QuotientsFinding limits by L'Hospital's RuleL'Hôpital's rule does not apply?!L'Hospital's rule helpCompute the limit without L'Hospital's ruleProof of L'Hospital's Rule
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Can we apply L'Hospital's rule?
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Can we apply L'Hospital's rule?
The 2019 Stack Overflow Developer Survey Results Are InResilient L'Hospital's rule questionUsing L'Hospital's Rule to evaluate limit to infinityIs 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 QuotientsFinding limits by L'Hospital's RuleL'Hôpital's rule does not apply?!L'Hospital's rule helpCompute 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 it's derivative may not be continuous at some point. One of many counter examples is:
$$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
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
$endgroup$
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:
$$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
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
$endgroup$
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:
$$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
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
$endgroup$
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:
$$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
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
1088
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.
$endgroup$
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ 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 $lim_hto 0f'(a+h)$ 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
40 mins ago
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ 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 $f'(a)$.
New contributor
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
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votes
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.
$endgroup$
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ 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 $lim_hto 0f'(a+h)$ 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
40 mins ago
add a comment |
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.
$endgroup$
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ 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 $lim_hto 0f'(a+h)$ 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
40 mins ago
add a comment |
$begingroup$
L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.
$endgroup$
L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.
answered 1 hour ago
HelmutHelmut
739128
739128
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ 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 $lim_hto 0f'(a+h)$ 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
40 mins ago
add a comment |
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ 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 $lim_hto 0f'(a+h)$ 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
40 mins ago
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ is one of $pminfty$, which some people classify as not existing, the implication also holds.
$endgroup$
– user647486
1 hour ago
$begingroup$
@Dhvanit when $lim_hto0fracf'(h)g'(h)$ 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 $lim_hto 0f'(a+h)$ 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
40 mins ago
$begingroup$
Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $lim_hto 0f'(a+h)$ 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
40 mins ago
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ 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 $f'(a)$.
New contributor
$endgroup$
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ 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 $f'(a)$.
New contributor
$endgroup$
add a comment |
$begingroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ 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 $f'(a)$.
New contributor
$endgroup$
In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ 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 $f'(a)$.
New contributor
New contributor
answered 1 hour ago
mihaildmihaild
5689
5689
New contributor
New contributor
add a comment |
add a comment |
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