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Incremental quotient limit
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If $f:mathbbRrightarrow mathbbR$ is differentiable, then $lim_hto 0dfracf(3+2h)-f(3)h=?$
I wanted to know if you could help me with this exercise, I know that you are using the incremental quotient.
The book's solution:
Here you have to think a little. They are giving us a limit that is very similar to the derivative by definition at point $x_0 = 3$, but it is exactly this derivative, because it would be like this:
$$f'(x_0)=lim_hto 0dfracf(x_0+h)-f(x_0)hRightarrow f'(3)=lim_hto 0dfracf(3+h)-f(3)h$$
As in the limit $f(3)$ appears, we can write it as $kf'(3)$, and that name we give to the exercise limit.
Compare $f'(3)$ and $kf'(3)$ using the definition and the limit they give us: (if they are equal, $k = 1$, and we are left with $f '(3)$, otherwise, it will be some other solution)
$$f'(3)=kf'(3)Rightarrow lim_hto 0dfracf(3+h)-f(3)h=lim_hto 0dfracf(3+2h)-f(3)hRightarrow lim_hto 03+h=lim_hto 03+2h$$
We have two limits that tend to $0$. We replace and we have left:
$$lim_hto 03+h=lim_hto 03+2hRightarrow 3+0=3+2cdot 0Rightarrow 3=3
$$
Then, the limits are equal, so $k = 1$. In this way, the answer is $f '(3)$.
real-analysis derivatives
edited Jul 31 at 16:40
John Ma
37.5k93669
asked Jul 29 at 16:24
a5537539
113
add a comment |Â
up vote
1
down vote
favorite
If $f:mathbbRrightarrow mathbbR$ is differentiable, then $lim_hto 0dfracf(3+2h)-f(3)h=?$
I wanted to know if you could help me with this exercise, I know that you are using the incremental quotient.
The book's solution:
Here you have to think a little. They are giving us a limit that is very similar to the derivative by definition at point $x_0 = 3$, but it is exactly this derivative, because it would be like this:
$$f'(x_0)=lim_hto 0dfracf(x_0+h)-f(x_0)hRightarrow f'(3)=lim_hto 0dfracf(3+h)-f(3)h$$
As in the limit $f(3)$ appears, we can write it as $kf'(3)$, and that name we give to the exercise limit.
Compare $f'(3)$ and $kf'(3)$ using the definition and the limit they give us: (if they are equal, $k = 1$, and we are left with $f '(3)$, otherwise, it will be some other solution)
$$f'(3)=kf'(3)Rightarrow lim_hto 0dfracf(3+h)-f(3)h=lim_hto 0dfracf(3+2h)-f(3)hRightarrow lim_hto 03+h=lim_hto 03+2h$$
We have two limits that tend to $0$. We replace and we have left:
$$lim_hto 03+h=lim_hto 03+2hRightarrow 3+0=3+2cdot 0Rightarrow 3=3
$$
Then, the limits are equal, so $k = 1$. In this way, the answer is $f '(3)$.
real-analysis derivatives
edited Jul 31 at 16:40
John Ma
37.5k93669
asked Jul 29 at 16:24
a5537539
113
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 29 at 16:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $f:mathbbRrightarrow mathbbR$ is differentiable, then $lim_hto 0dfracf(3+2h)-f(3)h=?$
I wanted to know if you could help me with this exercise, I know that you are using the incremental quotient.
The book's solution:
Here you have to think a little. They are giving us a limit that is very similar to the derivative by definition at point $x_0 = 3$, but it is exactly this derivative, because it would be like this:
$$f'(x_0)=lim_hto 0dfracf(x_0+h)-f(x_0)hRightarrow f'(3)=lim_hto 0dfracf(3+h)-f(3)h$$
As in the limit $f(3)$ appears, we can write it as $kf'(3)$, and that name we give to the exercise limit.
Compare $f'(3)$ and $kf'(3)$ using the definition and the limit they give us: (if they are equal, $k = 1$, and we are left with $f '(3)$, otherwise, it will be some other solution)
$$f'(3)=kf'(3)Rightarrow lim_hto 0dfracf(3+h)-f(3)h=lim_hto 0dfracf(3+2h)-f(3)hRightarrow lim_hto 03+h=lim_hto 03+2h$$
We have two limits that tend to $0$. We replace and we have left:
$$lim_hto 03+h=lim_hto 03+2hRightarrow 3+0=3+2cdot 0Rightarrow 3=3
$$
Then, the limits are equal, so $k = 1$. In this way, the answer is $f '(3)$.
real-analysis derivatives
edited Jul 31 at 16:40
John Ma
37.5k93669
asked Jul 29 at 16:24
a5537539
113
If $f:mathbbRrightarrow mathbbR$ is differentiable, then $lim_hto 0dfracf(3+2h)-f(3)h=?$
I wanted to know if you could help me with this exercise, I know that you are using the incremental quotient.
The book's solution:
Here you have to think a little. They are giving us a limit that is very similar to the derivative by definition at point $x_0 = 3$, but it is exactly this derivative, because it would be like this:
$$f'(x_0)=lim_hto 0dfracf(x_0+h)-f(x_0)hRightarrow f'(3)=lim_hto 0dfracf(3+h)-f(3)h$$
As in the limit $f(3)$ appears, we can write it as $kf'(3)$, and that name we give to the exercise limit.
Compare $f'(3)$ and $kf'(3)$ using the definition and the limit they give us: (if they are equal, $k = 1$, and we are left with $f '(3)$, otherwise, it will be some other solution)
$$f'(3)=kf'(3)Rightarrow lim_hto 0dfracf(3+h)-f(3)h=lim_hto 0dfracf(3+2h)-f(3)hRightarrow lim_hto 03+h=lim_hto 03+2h$$
We have two limits that tend to $0$. We replace and we have left:
$$lim_hto 03+h=lim_hto 03+2hRightarrow 3+0=3+2cdot 0Rightarrow 3=3
$$
Then, the limits are equal, so $k = 1$. In this way, the answer is $f '(3)$.
real-analysis derivatives
edited Jul 31 at 16:40
John Ma
37.5k93669
asked Jul 29 at 16:24
a5537539
113
edited Jul 31 at 16:40
John Ma
37.5k93669
edited Jul 31 at 16:40
John Ma
37.5k93669
edited Jul 31 at 16:40
John Ma
37.5k93669
37.5k93669
asked Jul 29 at 16:24
a5537539
113
asked Jul 29 at 16:24
a5537539
113
asked Jul 29 at 16:24
a5537539
113
113
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 29 at 16:28
add a comment |Â
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 29 at 16:28
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 29 at 16:28
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 29 at 16:28
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Big Hint: $$fracf(3+2h)-f(3)h=2cdotfracf(3+2h)-f(3)2h,$$ and $hto 0$ if and only if $2hto0.$
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
 |Â
show 5 more comments
up vote
1
down vote
Hint
What is $$limlimits_h to 0 fracf(3+2h)-f(3)2h?$$
What can you conclude from this limit?
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
add a comment |Â
up vote
0
down vote
$$lim_hto 0dfracf(3+2h)-f(3)h= lim_2hto 0 2dfracf(3+2h)-f(3)2h = 2f'(3) $$
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Big Hint: $$fracf(3+2h)-f(3)h=2cdotfracf(3+2h)-f(3)2h,$$ and $hto 0$ if and only if $2hto0.$
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
 |Â
show 5 more comments
up vote
2
down vote
accepted
Big Hint: $$fracf(3+2h)-f(3)h=2cdotfracf(3+2h)-f(3)2h,$$ and $hto 0$ if and only if $2hto0.$
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
 |Â
show 5 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Big Hint: $$fracf(3+2h)-f(3)h=2cdotfracf(3+2h)-f(3)2h,$$ and $hto 0$ if and only if $2hto0.$
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
Big Hint: $$fracf(3+2h)-f(3)h=2cdotfracf(3+2h)-f(3)2h,$$ and $hto 0$ if and only if $2hto0.$
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
answered Jul 29 at 16:26
Cameron Buie
83.5k771152
83.5k771152
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
 |Â
show 5 more comments
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
Thanks! So, this limit is equal to $2cdot f'(3)$, right?
– a5537539
Jul 29 at 16:31
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Bingo! Can you justify it?
– Cameron Buie
Jul 29 at 16:37
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Well, if $f'(3) happens to be $0,$ then that's true, but otherwise the book's answer is not correct.
– Cameron Buie
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
Then I do $t=2h$ and $hto 0$ if and only if $tto 0$, and I will have $2lim_tto 0dfracf(3+t)-f(3)t=2f'(3)$
– a5537539
Jul 29 at 16:38
 |Â
show 5 more comments
up vote
1
down vote
Hint
What is $$limlimits_h to 0 fracf(3+2h)-f(3)2h?$$
What can you conclude from this limit?
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
add a comment |Â
up vote
1
down vote
Hint
What is $$limlimits_h to 0 fracf(3+2h)-f(3)2h?$$
What can you conclude from this limit?
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint
What is $$limlimits_h to 0 fracf(3+2h)-f(3)2h?$$
What can you conclude from this limit?
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
Hint
What is $$limlimits_h to 0 fracf(3+2h)-f(3)2h?$$
What can you conclude from this limit?
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
answered Jul 29 at 16:27
mathcounterexamples.net
23.3k21651
23.3k21651
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
add a comment |Â
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
In a book, I have seen that the good answer is $f'(3)$. I don't know if it is well solved.
– a5537539
Jul 29 at 16:37
add a comment |Â
up vote
0
down vote
$$lim_hto 0dfracf(3+2h)-f(3)h= lim_2hto 0 2dfracf(3+2h)-f(3)2h = 2f'(3) $$
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
$$lim_hto 0dfracf(3+2h)-f(3)h= lim_2hto 0 2dfracf(3+2h)-f(3)2h = 2f'(3) $$
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$lim_hto 0dfracf(3+2h)-f(3)h= lim_2hto 0 2dfracf(3+2h)-f(3)2h = 2f'(3) $$
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
$$lim_hto 0dfracf(3+2h)-f(3)h= lim_2hto 0 2dfracf(3+2h)-f(3)2h = 2f'(3) $$
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
answered Jul 29 at 16:57
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
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Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Jul 29 at 16:28