Mixing
Definition of derivative - is it approaching value or actual value?
0
$begingroup$
When we say a limit,
say for example: $lim_limits{xto 2} f(x)$, where $f(x) = x^2$.
And we say this that as $x$ tends to $2$, the value of $f(x)$ approaches to $4$.
(I.e., the actual value of the function as $x$ tends to $2$ might not be $4$, but it surely is approaching to $4$ from either side.)
Then, we used this definition of limit to define derivative of a function at a point:
$$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
But while saying the value of derivative at some $x$, we say absolutely the value, not approaching.
For example, we say the derivative of $x^2$ at $x=3$ is $6$.
Shouldn't it be said that the derivative of $x^2$ at $x=3$ approaches $6$?
calculus limits derivatives definition
edited Jan 13 at 14:51
KM101
6,0151524
asked Jan 13 at 13:59
lancerlancer
154
$endgroup$
add a comment |
0
$begingroup$
When we say a limit,
say for example: $lim_limits{xto 2} f(x)$, where $f(x) = x^2$.
And we say this that as $x$ tends to $2$, the value of $f(x)$ approaches to $4$.
(I.e., the actual value of the function as $x$ tends to $2$ might not be $4$, but it surely is approaching to $4$ from either side.)
Then, we used this definition of limit to define derivative of a function at a point:
$$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
But while saying the value of derivative at some $x$, we say absolutely the value, not approaching.
For example, we say the derivative of $x^2$ at $x=3$ is $6$.
Shouldn't it be said that the derivative of $x^2$ at $x=3$ approaches $6$?
calculus limits derivatives definition
edited Jan 13 at 14:51
KM101
6,0151524
asked Jan 13 at 13:59
lancerlancer
154
$endgroup$
1
$begingroup$
The slope of the tangent at $x = 3$ is exactly $6$.
$endgroup$
– KM101
Jan 13 at 14:06
$begingroup$
Given a function $f$ and a number $a$ we can find the number $f(a) $ under certain conditions (condition being that $a$ is in domain of $f$). In a similar manner given function $f$ and a number $a$ we can talk of the number $lim_{xto a} f(x) $ under certain more stringent conditions. Understanding those conditions is the key to understanding limits.
$endgroup$
– Paramanand Singh
Jan 13 at 16:55
add a comment |
0
0
0
$begingroup$
When we say a limit,
say for example: $lim_limits{xto 2} f(x)$, where $f(x) = x^2$.
And we say this that as $x$ tends to $2$, the value of $f(x)$ approaches to $4$.
(I.e., the actual value of the function as $x$ tends to $2$ might not be $4$, but it surely is approaching to $4$ from either side.)
Then, we used this definition of limit to define derivative of a function at a point:
$$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
But while saying the value of derivative at some $x$, we say absolutely the value, not approaching.
For example, we say the derivative of $x^2$ at $x=3$ is $6$.
Shouldn't it be said that the derivative of $x^2$ at $x=3$ approaches $6$?
calculus limits derivatives definition
edited Jan 13 at 14:51
KM101
6,0151524
asked Jan 13 at 13:59
lancerlancer
154
$endgroup$
When we say a limit,
say for example: $lim_limits{xto 2} f(x)$, where $f(x) = x^2$.
And we say this that as $x$ tends to $2$, the value of $f(x)$ approaches to $4$.
(I.e., the actual value of the function as $x$ tends to $2$ might not be $4$, but it surely is approaching to $4$ from either side.)
Then, we used this definition of limit to define derivative of a function at a point:
$$lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
But while saying the value of derivative at some $x$, we say absolutely the value, not approaching.
For example, we say the derivative of $x^2$ at $x=3$ is $6$.
Shouldn't it be said that the derivative of $x^2$ at $x=3$ approaches $6$?
calculus limits derivatives definition
calculus limits derivatives definition
edited Jan 13 at 14:51
KM101
6,0151524
asked Jan 13 at 13:59
lancerlancer
154
edited Jan 13 at 14:51
KM101
6,0151524
asked Jan 13 at 13:59
lancerlancer
154
edited Jan 13 at 14:51
KM101
6,0151524
edited Jan 13 at 14:51
KM101
6,0151524
edited Jan 13 at 14:51
KM101
6,0151524
6,0151524
asked Jan 13 at 13:59
lancerlancer
154
asked Jan 13 at 13:59
lancerlancer
154
asked Jan 13 at 13:59
lancerlancer
154
154
1
$begingroup$
The slope of the tangent at $x = 3$ is exactly $6$.
$endgroup$
– KM101
Jan 13 at 14:06
$begingroup$
Given a function $f$ and a number $a$ we can find the number $f(a) $ under certain conditions (condition being that $a$ is in domain of $f$). In a similar manner given function $f$ and a number $a$ we can talk of the number $lim_{xto a} f(x) $ under certain more stringent conditions. Understanding those conditions is the key to understanding limits.
$endgroup$
– Paramanand Singh
Jan 13 at 16:55
add a comment |
1
$begingroup$
The slope of the tangent at $x = 3$ is exactly $6$.
$endgroup$
– KM101
Jan 13 at 14:06
$begingroup$
Given a function $f$ and a number $a$ we can find the number $f(a) $ under certain conditions (condition being that $a$ is in domain of $f$). In a similar manner given function $f$ and a number $a$ we can talk of the number $lim_{xto a} f(x) $ under certain more stringent conditions. Understanding those conditions is the key to understanding limits.
$endgroup$
– Paramanand Singh
Jan 13 at 16:55
1
1
$begingroup$
The slope of the tangent at $x = 3$ is exactly $6$.
$endgroup$
– KM101
Jan 13 at 14:06
$begingroup$
The slope of the tangent at $x = 3$ is exactly $6$.
$endgroup$
– KM101
Jan 13 at 14:06
$begingroup$
Given a function $f$ and a number $a$ we can find the number $f(a) $ under certain conditions (condition being that $a$ is in domain of $f$). In a similar manner given function $f$ and a number $a$ we can talk of the number $lim_{xto a} f(x) $ under certain more stringent conditions. Understanding those conditions is the key to understanding limits.
$endgroup$
– Paramanand Singh
Jan 13 at 16:55
$begingroup$
Given a function $f$ and a number $a$ we can find the number $f(a) $ under certain conditions (condition being that $a$ is in domain of $f$). In a similar manner given function $f$ and a number $a$ we can talk of the number $lim_{xto a} f(x) $ under certain more stringent conditions. Understanding those conditions is the key to understanding limits.
$endgroup$
– Paramanand Singh
Jan 13 at 16:55
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
"Approaches" is more of colloquial term. There is a rigorous definition of the limit, and it says that the limit is (exactly!) equal to something (If the limit exists).
The derivative of $x^2$ at $x=3$ is equal to $6$, not approaching $6$.
answered Jan 13 at 14:19
EffEff
11.6k21638
$endgroup$
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
|
show 7 more comments
1
$begingroup$
Let $f(x)=x+1$. So $lim_{xto 2}f(x)=3$. That's the same as saying "the limit of $f(x)$ as $x$ approaches $2$ is $3$". It's also the same as saying "$f(x)$ approaches $3$ as $x$ approaches $2$".
But not "the limit of $f(x)$ approaches $3$"!!! Students say that sometimes - they shouldn't. That limit is a number. One single number, doesn't "approach" anything, in fact it equals $3$.
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
"Approaches" is more of colloquial term. There is a rigorous definition of the limit, and it says that the limit is (exactly!) equal to something (If the limit exists).
The derivative of $x^2$ at $x=3$ is equal to $6$, not approaching $6$.
answered Jan 13 at 14:19
EffEff
11.6k21638
$endgroup$
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
|
show 7 more comments
1
$begingroup$
"Approaches" is more of colloquial term. There is a rigorous definition of the limit, and it says that the limit is (exactly!) equal to something (If the limit exists).
The derivative of $x^2$ at $x=3$ is equal to $6$, not approaching $6$.
answered Jan 13 at 14:19
EffEff
11.6k21638
$endgroup$
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
|
show 7 more comments
1
1
1
$begingroup$
"Approaches" is more of colloquial term. There is a rigorous definition of the limit, and it says that the limit is (exactly!) equal to something (If the limit exists).
The derivative of $x^2$ at $x=3$ is equal to $6$, not approaching $6$.
answered Jan 13 at 14:19
EffEff
11.6k21638
$endgroup$
"Approaches" is more of colloquial term. There is a rigorous definition of the limit, and it says that the limit is (exactly!) equal to something (If the limit exists).
The derivative of $x^2$ at $x=3$ is equal to $6$, not approaching $6$.
answered Jan 13 at 14:19
EffEff
11.6k21638
answered Jan 13 at 14:19
EffEff
11.6k21638
answered Jan 13 at 14:19
EffEff
11.6k21638
answered Jan 13 at 14:19
EffEff
11.6k21638
11.6k21638
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
|
show 7 more comments
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
Thanks Eff for the comment. I know that the value of derivative has to be a value, not approaching value. But if you see in more detail terms, we need minimum of two points on a curve to define a line. And to define tangent on a curve at any point, we chose {x, f(x)} and lim h-->0 { x+h, f(x+h) } as the second point. Now with these two points in hand, we defined the tangent at that point. But technically speaking, even though this second point is very very close to f(x), still its a different point.
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
{Continuing from above} So still shouldnt we be getting a very very close (but yet approximate value) of the slope of the tangent?
$endgroup$
– lancer
Jan 14 at 6:20
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
I would also like to retype the comment I gave above in one of the answers for your reference. Derivative of x^2 Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:49
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
You know the definition of the derivative using first principles. But do you know the the rigorous definition of a limit (which is needed for derivatives)? Your questions makes it appear (to me) that this is where the confuses arises.
$endgroup$
– Eff
Jan 14 at 9:00
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
$begingroup$
Kindly elucidate what you mean.
$endgroup$
– lancer
Jan 14 at 14:09
|
show 7 more comments
1
$begingroup$
Let $f(x)=x+1$. So $lim_{xto 2}f(x)=3$. That's the same as saying "the limit of $f(x)$ as $x$ approaches $2$ is $3$". It's also the same as saying "$f(x)$ approaches $3$ as $x$ approaches $2$".
But not "the limit of $f(x)$ approaches $3$"!!! Students say that sometimes - they shouldn't. That limit is a number. One single number, doesn't "approach" anything, in fact it equals $3$.
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
add a comment |
1
$begingroup$
Let $f(x)=x+1$. So $lim_{xto 2}f(x)=3$. That's the same as saying "the limit of $f(x)$ as $x$ approaches $2$ is $3$". It's also the same as saying "$f(x)$ approaches $3$ as $x$ approaches $2$".
But not "the limit of $f(x)$ approaches $3$"!!! Students say that sometimes - they shouldn't. That limit is a number. One single number, doesn't "approach" anything, in fact it equals $3$.
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
add a comment |
1
1
1
$begingroup$
Let $f(x)=x+1$. So $lim_{xto 2}f(x)=3$. That's the same as saying "the limit of $f(x)$ as $x$ approaches $2$ is $3$". It's also the same as saying "$f(x)$ approaches $3$ as $x$ approaches $2$".
But not "the limit of $f(x)$ approaches $3$"!!! Students say that sometimes - they shouldn't. That limit is a number. One single number, doesn't "approach" anything, in fact it equals $3$.
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
$endgroup$
Let $f(x)=x+1$. So $lim_{xto 2}f(x)=3$. That's the same as saying "the limit of $f(x)$ as $x$ approaches $2$ is $3$". It's also the same as saying "$f(x)$ approaches $3$ as $x$ approaches $2$".
But not "the limit of $f(x)$ approaches $3$"!!! Students say that sometimes - they shouldn't. That limit is a number. One single number, doesn't "approach" anything, in fact it equals $3$.
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
answered Jan 13 at 14:48
David C. UllrichDavid C. Ullrich
60.9k43994
60.9k43994
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
add a comment |
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
+1 I think the word approches is a misnomer and culprit here as it leads to the idea that some sort of motion/action is happening. Or perhaps a process of assigning values to $x$ and getting corresponding values of $f(x) $ is in place.
$endgroup$
– Paramanand Singh
Jan 13 at 16:52
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
Thanks Mr. Ullrich and Mr Singh. This line clears the limit for me probably - "f(x) approaches 3 as x approaches 2" and therefore, the value of limit as x approaches 2 is equal to 3. However, if I want to find the value of the derivative of the function at x=2, I would need one more point which is very very near to x=2 (and not x=2, only near). In that case if the actual value of limit lim x-->2 is equal to 3, shouldnt, the value of limh→0 [f(x+h)−f(x) ](h) comes out as {3-3/0} ? I know how to solve for derivatives and limits, but its this concept that I am failing to understand.
$endgroup$
– lancer
Jan 14 at 6:13
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
$begingroup$
I would also like to highlight the derivation for the derivative of x^2. Using the first principle of derivative, lim h-->0 [{ (x+h)^2 - x^2 }/h] we get in the end as lim h--> 0 { h + 2x } , and then we put the value if h -->0 and get the derivative as 2x. My point is, just here, havnt we made our derivative inducive of an infinitesimal error? As the value could have neem 2x + some infinitesimal value. This error may not be significant but doesnt it make the whole derivative and approximate or approaching value?
$endgroup$
– lancer
Jan 14 at 6:48
add a comment |
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$begingroup$
The slope of the tangent at $x = 3$ is exactly $6$.
$endgroup$
– KM101
Jan 13 at 14:06
$begingroup$
Given a function $f$ and a number $a$ we can find the number $f(a) $ under certain conditions (condition being that $a$ is in domain of $f$). In a similar manner given function $f$ and a number $a$ we can talk of the number $lim_{xto a} f(x) $ under certain more stringent conditions. Understanding those conditions is the key to understanding limits.
$endgroup$
– Paramanand Singh
Jan 13 at 16:55