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Why can't we apply the L'Hopital rule once the function is no longer an indeterminate form?
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As I understand it, L'Hopital Rule comes from Cauchy's Generalised Mean Value Theorem, which states that if 2 functions, $f(x)$ and $g(x)$ are continuous and differentiable over the interval $[a,b]$ and $(a,b)$ respectively, then $∃$
$c∈[a,b]$ such that
$$(f(b)-f(a))/(g(b)-g(a))=(f'(c))/(g'(c))$$
Now if $f(x)$ and $g(x)$ are arbitrary functions, why shouldn't L'Hopital Rule hold once the function is no longer indeterminate? Is there something I didn't see in the proof?
calculus
asked yesterday
Yip Jung Hon
14511
 |Â
show 1 more comment
up vote
2
down vote
favorite
As I understand it, L'Hopital Rule comes from Cauchy's Generalised Mean Value Theorem, which states that if 2 functions, $f(x)$ and $g(x)$ are continuous and differentiable over the interval $[a,b]$ and $(a,b)$ respectively, then $∃$
$c∈[a,b]$ such that
$$(f(b)-f(a))/(g(b)-g(a))=(f'(c))/(g'(c))$$
Now if $f(x)$ and $g(x)$ are arbitrary functions, why shouldn't L'Hopital Rule hold once the function is no longer indeterminate? Is there something I didn't see in the proof?
calculus
asked yesterday
Yip Jung Hon
14511
Do you mean that $bto a$ (to ensure that $cto a$)?
– A.Γ.
yesterday
1
Consider the limit of $frac3x^22x$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer!
– Sufaid Saleel
yesterday
@A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies.
– Yip Jung Hon
yesterday
@SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate.
– Yip Jung Hon
yesterday
@YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question.
– A.Γ.
yesterday
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
As I understand it, L'Hopital Rule comes from Cauchy's Generalised Mean Value Theorem, which states that if 2 functions, $f(x)$ and $g(x)$ are continuous and differentiable over the interval $[a,b]$ and $(a,b)$ respectively, then $∃$
$c∈[a,b]$ such that
$$(f(b)-f(a))/(g(b)-g(a))=(f'(c))/(g'(c))$$
Now if $f(x)$ and $g(x)$ are arbitrary functions, why shouldn't L'Hopital Rule hold once the function is no longer indeterminate? Is there something I didn't see in the proof?
calculus
asked yesterday
Yip Jung Hon
14511
As I understand it, L'Hopital Rule comes from Cauchy's Generalised Mean Value Theorem, which states that if 2 functions, $f(x)$ and $g(x)$ are continuous and differentiable over the interval $[a,b]$ and $(a,b)$ respectively, then $∃$
$c∈[a,b]$ such that
$$(f(b)-f(a))/(g(b)-g(a))=(f'(c))/(g'(c))$$
Now if $f(x)$ and $g(x)$ are arbitrary functions, why shouldn't L'Hopital Rule hold once the function is no longer indeterminate? Is there something I didn't see in the proof?
calculus
asked yesterday
Yip Jung Hon
14511
asked yesterday
Yip Jung Hon
14511
asked yesterday
Yip Jung Hon
14511
asked yesterday
Yip Jung Hon
14511
14511
Do you mean that $bto a$ (to ensure that $cto a$)?
– A.Γ.
yesterday
1
Consider the limit of $frac3x^22x$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer!
– Sufaid Saleel
yesterday
@A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies.
– Yip Jung Hon
yesterday
@SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate.
– Yip Jung Hon
yesterday
@YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question.
– A.Γ.
yesterday
 |Â
show 1 more comment
Do you mean that $bto a$ (to ensure that $cto a$)?
– A.Γ.
yesterday
1
Consider the limit of $frac3x^22x$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer!
– Sufaid Saleel
yesterday
@A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies.
– Yip Jung Hon
yesterday
@SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate.
– Yip Jung Hon
yesterday
@YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question.
– A.Γ.
yesterday
Do you mean that $bto a$ (to ensure that $cto a$)?
– A.Γ.
yesterday
Do you mean that $bto a$ (to ensure that $cto a$)?
– A.Γ.
yesterday
1
1
Consider the limit of $frac3x^22x$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer!
– Sufaid Saleel
yesterday
Consider the limit of $frac3x^22x$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer!
– Sufaid Saleel
yesterday
@A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies.
– Yip Jung Hon
yesterday
@A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies.
– Yip Jung Hon
yesterday
@SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate.
– Yip Jung Hon
yesterday
@SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate.
– Yip Jung Hon
yesterday
@YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question.
– A.Γ.
yesterday
@YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question.
– A.Γ.
yesterday
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
When $lim_xto af(x)=lim_xto ag(x)=0$, then we can apply Cauchy's theorem:$$fracf(x)g(x)=fracf(x)-f(a)g(x)-f(a)=fracf'(c)g'(c),$$for some $c$ and therefore$$lim_xto afracf'(x)g'(x)=Limplieslim_xto afracf(x)g(x)=L.$$But if $lim_xto af(x)neq0$ or $lim_xto ag(x)neq0$, this argument doesn't apply anymore.
Of course, this doesn't actually prove that we can't apply L'Hopital's rule then. What proves that is the existence of counter examples.
answered yesterday
José Carlos Santos
111k1695171
add a comment |Â
up vote
5
down vote
I usually motivate L'hospital with Taylor series (or linear approximation if it's early in the course.) See that (using $a=0$ for convenience)
$$lim_xto 0 fracf(x)g(x) = lim_xto 0 fracf(0) +f'(0)x+f''(0)x^2/2+cdotsg(0)+g'(0)x+g''(0)x^2/2+cdots.$$
If both $f(0)=0$ and $g(0)=0$, this becomes
$$lim_xto 0 fracf'(0)x+f''(0)x^2/2+cdotsg'(0)x+g''(0)x^2/2+cdots.$$
But if they are not both $0$, then I can't do this step. Then I can cancel one $x$ in the fraction to get
$$lim_xto 0 fracf'(0)+f''(0)x/2+cdotsg'(0)+g''(0)x/2+cdots,$$
and we can repeat this reasoning until we hit a derivative that's not $0$ at $0$.
answered 21 hours ago
B. Goddard
16.4k21238
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
When $lim_xto af(x)=lim_xto ag(x)=0$, then we can apply Cauchy's theorem:$$fracf(x)g(x)=fracf(x)-f(a)g(x)-f(a)=fracf'(c)g'(c),$$for some $c$ and therefore$$lim_xto afracf'(x)g'(x)=Limplieslim_xto afracf(x)g(x)=L.$$But if $lim_xto af(x)neq0$ or $lim_xto ag(x)neq0$, this argument doesn't apply anymore.
Of course, this doesn't actually prove that we can't apply L'Hopital's rule then. What proves that is the existence of counter examples.
answered yesterday
José Carlos Santos
111k1695171
add a comment |Â
up vote
4
down vote
accepted
When $lim_xto af(x)=lim_xto ag(x)=0$, then we can apply Cauchy's theorem:$$fracf(x)g(x)=fracf(x)-f(a)g(x)-f(a)=fracf'(c)g'(c),$$for some $c$ and therefore$$lim_xto afracf'(x)g'(x)=Limplieslim_xto afracf(x)g(x)=L.$$But if $lim_xto af(x)neq0$ or $lim_xto ag(x)neq0$, this argument doesn't apply anymore.
Of course, this doesn't actually prove that we can't apply L'Hopital's rule then. What proves that is the existence of counter examples.
answered yesterday
José Carlos Santos
111k1695171
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
When $lim_xto af(x)=lim_xto ag(x)=0$, then we can apply Cauchy's theorem:$$fracf(x)g(x)=fracf(x)-f(a)g(x)-f(a)=fracf'(c)g'(c),$$for some $c$ and therefore$$lim_xto afracf'(x)g'(x)=Limplieslim_xto afracf(x)g(x)=L.$$But if $lim_xto af(x)neq0$ or $lim_xto ag(x)neq0$, this argument doesn't apply anymore.
Of course, this doesn't actually prove that we can't apply L'Hopital's rule then. What proves that is the existence of counter examples.
answered yesterday
José Carlos Santos
111k1695171
When $lim_xto af(x)=lim_xto ag(x)=0$, then we can apply Cauchy's theorem:$$fracf(x)g(x)=fracf(x)-f(a)g(x)-f(a)=fracf'(c)g'(c),$$for some $c$ and therefore$$lim_xto afracf'(x)g'(x)=Limplieslim_xto afracf(x)g(x)=L.$$But if $lim_xto af(x)neq0$ or $lim_xto ag(x)neq0$, this argument doesn't apply anymore.
Of course, this doesn't actually prove that we can't apply L'Hopital's rule then. What proves that is the existence of counter examples.
answered yesterday
José Carlos Santos
111k1695171
answered yesterday
José Carlos Santos
111k1695171
answered yesterday
José Carlos Santos
111k1695171
answered yesterday
José Carlos Santos
111k1695171
111k1695171
add a comment |Â
add a comment |Â
up vote
5
down vote
I usually motivate L'hospital with Taylor series (or linear approximation if it's early in the course.) See that (using $a=0$ for convenience)
$$lim_xto 0 fracf(x)g(x) = lim_xto 0 fracf(0) +f'(0)x+f''(0)x^2/2+cdotsg(0)+g'(0)x+g''(0)x^2/2+cdots.$$
If both $f(0)=0$ and $g(0)=0$, this becomes
$$lim_xto 0 fracf'(0)x+f''(0)x^2/2+cdotsg'(0)x+g''(0)x^2/2+cdots.$$
But if they are not both $0$, then I can't do this step. Then I can cancel one $x$ in the fraction to get
$$lim_xto 0 fracf'(0)+f''(0)x/2+cdotsg'(0)+g''(0)x/2+cdots,$$
and we can repeat this reasoning until we hit a derivative that's not $0$ at $0$.
answered 21 hours ago
B. Goddard
16.4k21238
add a comment |Â
up vote
5
down vote
I usually motivate L'hospital with Taylor series (or linear approximation if it's early in the course.) See that (using $a=0$ for convenience)
$$lim_xto 0 fracf(x)g(x) = lim_xto 0 fracf(0) +f'(0)x+f''(0)x^2/2+cdotsg(0)+g'(0)x+g''(0)x^2/2+cdots.$$
If both $f(0)=0$ and $g(0)=0$, this becomes
$$lim_xto 0 fracf'(0)x+f''(0)x^2/2+cdotsg'(0)x+g''(0)x^2/2+cdots.$$
But if they are not both $0$, then I can't do this step. Then I can cancel one $x$ in the fraction to get
$$lim_xto 0 fracf'(0)+f''(0)x/2+cdotsg'(0)+g''(0)x/2+cdots,$$
and we can repeat this reasoning until we hit a derivative that's not $0$ at $0$.
answered 21 hours ago
B. Goddard
16.4k21238
add a comment |Â
up vote
5
down vote
up vote
5
down vote
I usually motivate L'hospital with Taylor series (or linear approximation if it's early in the course.) See that (using $a=0$ for convenience)
$$lim_xto 0 fracf(x)g(x) = lim_xto 0 fracf(0) +f'(0)x+f''(0)x^2/2+cdotsg(0)+g'(0)x+g''(0)x^2/2+cdots.$$
If both $f(0)=0$ and $g(0)=0$, this becomes
$$lim_xto 0 fracf'(0)x+f''(0)x^2/2+cdotsg'(0)x+g''(0)x^2/2+cdots.$$
But if they are not both $0$, then I can't do this step. Then I can cancel one $x$ in the fraction to get
$$lim_xto 0 fracf'(0)+f''(0)x/2+cdotsg'(0)+g''(0)x/2+cdots,$$
and we can repeat this reasoning until we hit a derivative that's not $0$ at $0$.
answered 21 hours ago
B. Goddard
16.4k21238
I usually motivate L'hospital with Taylor series (or linear approximation if it's early in the course.) See that (using $a=0$ for convenience)
$$lim_xto 0 fracf(x)g(x) = lim_xto 0 fracf(0) +f'(0)x+f''(0)x^2/2+cdotsg(0)+g'(0)x+g''(0)x^2/2+cdots.$$
If both $f(0)=0$ and $g(0)=0$, this becomes
$$lim_xto 0 fracf'(0)x+f''(0)x^2/2+cdotsg'(0)x+g''(0)x^2/2+cdots.$$
But if they are not both $0$, then I can't do this step. Then I can cancel one $x$ in the fraction to get
$$lim_xto 0 fracf'(0)+f''(0)x/2+cdotsg'(0)+g''(0)x/2+cdots,$$
and we can repeat this reasoning until we hit a derivative that's not $0$ at $0$.
answered 21 hours ago
B. Goddard
16.4k21238
answered 21 hours ago
B. Goddard
16.4k21238
answered 21 hours ago
B. Goddard
16.4k21238
answered 21 hours ago
B. Goddard
16.4k21238
16.4k21238
add a comment |Â
add a comment |Â
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Do you mean that $bto a$ (to ensure that $cto a$)?
– A.Γ.
yesterday
1
Consider the limit of $frac3x^22x$ where x is tends to 1.Now putting $x=1$ we have the value $3/2$.But if we apply L'Hospital it will give us wrong answer!
– Sufaid Saleel
yesterday
@A.Γ. I'm not sure what you mean. From my understanding of Cauchy's Mean Value Theorem, a and b are boundaries you define for a function, and c is a point on the function where the relationship stated applies.
– Yip Jung Hon
yesterday
@SufaidSaleel Yes, that's true. But I don't understand why you have to stop applying L'Hopital Rule once the function is determinate.
– Yip Jung Hon
yesterday
@YipJungHon L'Hopital rule apples to limits. It is not clear what limit that you mean in the question.
– A.Γ.
yesterday