Is my conception of limit correct?
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$(1)$ Can I define the derivative
$left( dfrac{dy}{dx}=lim_{Delta x rightarrow 0}dfrac{Delta y}{Delta x} right)$ as a value which can never be reached when $Delta x$ approaches zero but every value smaller (or greater in other cases) to it can be reached when $Delta x$ approaches zero.
$(2) $Similarly can I define the definite integral
$displaystyle left( int^b_a y dx right)$
as the value of Reimann sum (in which we take the smallest value of function on the interval $Delta x$) which can never be reached when $Delta x$ approaches zero but every value smaller to it can be reached when $Delta x$ approaches zero.
calculus limits derivatives definite-integrals definition
asked Dec 9 '18 at 13:58
JoeJoe
292113
$endgroup$
add a comment |
$begingroup$
$(1)$ Can I define the derivative
$left( dfrac{dy}{dx}=lim_{Delta x rightarrow 0}dfrac{Delta y}{Delta x} right)$ as a value which can never be reached when $Delta x$ approaches zero but every value smaller (or greater in other cases) to it can be reached when $Delta x$ approaches zero.
$(2) $Similarly can I define the definite integral
$displaystyle left( int^b_a y dx right)$
as the value of Reimann sum (in which we take the smallest value of function on the interval $Delta x$) which can never be reached when $Delta x$ approaches zero but every value smaller to it can be reached when $Delta x$ approaches zero.
calculus limits derivatives definite-integrals definition
asked Dec 9 '18 at 13:58
JoeJoe
292113
$endgroup$
$begingroup$
Be sure not to confuse $Delta x$ and $x$. The definition of a derivative is $lim_limits{Delta xto 0}frac{Delta y}{Delta x}$.
$endgroup$
– KM101
Dec 9 '18 at 14:09
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Sorry......mistyping
$endgroup$
– Joe
Dec 9 '18 at 14:23
1
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No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
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– Christoph
Dec 9 '18 at 14:43
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@Christoph: Thanks.... I see.... So constant functions do not fit here. Any other flaws?
$endgroup$
– Joe
Dec 9 '18 at 14:47
add a comment |
$begingroup$
$(1)$ Can I define the derivative
$left( dfrac{dy}{dx}=lim_{Delta x rightarrow 0}dfrac{Delta y}{Delta x} right)$ as a value which can never be reached when $Delta x$ approaches zero but every value smaller (or greater in other cases) to it can be reached when $Delta x$ approaches zero.
$(2) $Similarly can I define the definite integral
$displaystyle left( int^b_a y dx right)$
as the value of Reimann sum (in which we take the smallest value of function on the interval $Delta x$) which can never be reached when $Delta x$ approaches zero but every value smaller to it can be reached when $Delta x$ approaches zero.
calculus limits derivatives definite-integrals definition
asked Dec 9 '18 at 13:58
JoeJoe
292113
$endgroup$
$(1)$ Can I define the derivative
$left( dfrac{dy}{dx}=lim_{Delta x rightarrow 0}dfrac{Delta y}{Delta x} right)$ as a value which can never be reached when $Delta x$ approaches zero but every value smaller (or greater in other cases) to it can be reached when $Delta x$ approaches zero.
$(2) $Similarly can I define the definite integral
$displaystyle left( int^b_a y dx right)$
as the value of Reimann sum (in which we take the smallest value of function on the interval $Delta x$) which can never be reached when $Delta x$ approaches zero but every value smaller to it can be reached when $Delta x$ approaches zero.
calculus limits derivatives definite-integrals definition
calculus limits derivatives definite-integrals definition
asked Dec 9 '18 at 13:58
JoeJoe
292113
asked Dec 9 '18 at 13:58
JoeJoe
292113
edited Dec 9 '18 at 14:33
Joe
asked Dec 9 '18 at 13:58
JoeJoe
292113
asked Dec 9 '18 at 13:58
JoeJoe
292113
asked Dec 9 '18 at 13:58
JoeJoe
292113
292113
$begingroup$
Be sure not to confuse $Delta x$ and $x$. The definition of a derivative is $lim_limits{Delta xto 0}frac{Delta y}{Delta x}$.
$endgroup$
– KM101
Dec 9 '18 at 14:09
$begingroup$
Sorry......mistyping
$endgroup$
– Joe
Dec 9 '18 at 14:23
1
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
$endgroup$
– Christoph
Dec 9 '18 at 14:43
$begingroup$
@Christoph: Thanks.... I see.... So constant functions do not fit here. Any other flaws?
$endgroup$
– Joe
Dec 9 '18 at 14:47
add a comment |
$begingroup$
Be sure not to confuse $Delta x$ and $x$. The definition of a derivative is $lim_limits{Delta xto 0}frac{Delta y}{Delta x}$.
$endgroup$
– KM101
Dec 9 '18 at 14:09
$begingroup$
Sorry......mistyping
$endgroup$
– Joe
Dec 9 '18 at 14:23
1
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
$endgroup$
– Christoph
Dec 9 '18 at 14:43
$begingroup$
@Christoph: Thanks.... I see.... So constant functions do not fit here. Any other flaws?
$endgroup$
– Joe
Dec 9 '18 at 14:47
$begingroup$
Be sure not to confuse $Delta x$ and $x$. The definition of a derivative is $lim_limits{Delta xto 0}frac{Delta y}{Delta x}$.
$endgroup$
– KM101
Dec 9 '18 at 14:09
$begingroup$
Be sure not to confuse $Delta x$ and $x$. The definition of a derivative is $lim_limits{Delta xto 0}frac{Delta y}{Delta x}$.
$endgroup$
– KM101
Dec 9 '18 at 14:09
$begingroup$
Sorry......mistyping
$endgroup$
– Joe
Dec 9 '18 at 14:23
$begingroup$
Sorry......mistyping
$endgroup$
– Joe
Dec 9 '18 at 14:23
1
1
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
$endgroup$
– Christoph
Dec 9 '18 at 14:43
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
$endgroup$
– Christoph
Dec 9 '18 at 14:43
$begingroup$
@Christoph: Thanks.... I see.... So constant functions do not fit here. Any other flaws?
$endgroup$
– Joe
Dec 9 '18 at 14:47
$begingroup$
@Christoph: Thanks.... I see.... So constant functions do not fit here. Any other flaws?
$endgroup$
– Joe
Dec 9 '18 at 14:47
add a comment |
2 Answers
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No. Both are false. The value might be reached (trivial example for the derivative: a linear function). And it is also false that "every" value ( in what range?) is attained, as the same example shows.
Your assertion is ( more or less) right if applied to the values of the variable , instead of to the function.
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
$endgroup$
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
add a comment |
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
The right description is that $lim_{Delta xto 0} frac{Delta y}{Delta x} = L$ means that for any choice of a small $varepsilon>0$ the values of $frac{Delta y}{Delta x}$ are inside the range $(L-varepsilon, L+varepsilon)$ as long as $Delta x$ is close enough to $0$. More precisely, for any $varepsilon >0$ there exists $delta > 0$ such that when $-delta < Delta x < delta$, then $L-varepsilon < frac{Delta y}{Delta x} < L+varepsilon$.
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
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add a comment |
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2 Answers
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$begingroup$
No. Both are false. The value might be reached (trivial example for the derivative: a linear function). And it is also false that "every" value ( in what range?) is attained, as the same example shows.
Your assertion is ( more or less) right if applied to the values of the variable , instead of to the function.
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
$endgroup$
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
add a comment |
$begingroup$
No. Both are false. The value might be reached (trivial example for the derivative: a linear function). And it is also false that "every" value ( in what range?) is attained, as the same example shows.
Your assertion is ( more or less) right if applied to the values of the variable , instead of to the function.
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
$endgroup$
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
add a comment |
$begingroup$
No. Both are false. The value might be reached (trivial example for the derivative: a linear function). And it is also false that "every" value ( in what range?) is attained, as the same example shows.
Your assertion is ( more or less) right if applied to the values of the variable , instead of to the function.
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
$endgroup$
No. Both are false. The value might be reached (trivial example for the derivative: a linear function). And it is also false that "every" value ( in what range?) is attained, as the same example shows.
Your assertion is ( more or less) right if applied to the values of the variable , instead of to the function.
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
answered Dec 9 '18 at 14:09
leonbloyleonbloy
41.1k645107
41.1k645107
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
add a comment |
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
$begingroup$
I have edited... Please have a look at it
$endgroup$
– Joe
Dec 9 '18 at 14:34
add a comment |
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
The right description is that $lim_{Delta xto 0} frac{Delta y}{Delta x} = L$ means that for any choice of a small $varepsilon>0$ the values of $frac{Delta y}{Delta x}$ are inside the range $(L-varepsilon, L+varepsilon)$ as long as $Delta x$ is close enough to $0$. More precisely, for any $varepsilon >0$ there exists $delta > 0$ such that when $-delta < Delta x < delta$, then $L-varepsilon < frac{Delta y}{Delta x} < L+varepsilon$.
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
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add a comment |
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
The right description is that $lim_{Delta xto 0} frac{Delta y}{Delta x} = L$ means that for any choice of a small $varepsilon>0$ the values of $frac{Delta y}{Delta x}$ are inside the range $(L-varepsilon, L+varepsilon)$ as long as $Delta x$ is close enough to $0$. More precisely, for any $varepsilon >0$ there exists $delta > 0$ such that when $-delta < Delta x < delta$, then $L-varepsilon < frac{Delta y}{Delta x} < L+varepsilon$.
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
$endgroup$
add a comment |
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
The right description is that $lim_{Delta xto 0} frac{Delta y}{Delta x} = L$ means that for any choice of a small $varepsilon>0$ the values of $frac{Delta y}{Delta x}$ are inside the range $(L-varepsilon, L+varepsilon)$ as long as $Delta x$ is close enough to $0$. More precisely, for any $varepsilon >0$ there exists $delta > 0$ such that when $-delta < Delta x < delta$, then $L-varepsilon < frac{Delta y}{Delta x} < L+varepsilon$.
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
$endgroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
The right description is that $lim_{Delta xto 0} frac{Delta y}{Delta x} = L$ means that for any choice of a small $varepsilon>0$ the values of $frac{Delta y}{Delta x}$ are inside the range $(L-varepsilon, L+varepsilon)$ as long as $Delta x$ is close enough to $0$. More precisely, for any $varepsilon >0$ there exists $delta > 0$ such that when $-delta < Delta x < delta$, then $L-varepsilon < frac{Delta y}{Delta x} < L+varepsilon$.
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
answered Dec 9 '18 at 14:47
ChristophChristoph
12.1k1642
12.1k1642
add a comment |
add a comment |
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$begingroup$
Be sure not to confuse $Delta x$ and $x$. The definition of a derivative is $lim_limits{Delta xto 0}frac{Delta y}{Delta x}$.
$endgroup$
– KM101
Dec 9 '18 at 14:09
$begingroup$
Sorry......mistyping
$endgroup$
– Joe
Dec 9 '18 at 14:23
1
$begingroup$
No. Consider the example where $y=1$ for all values of $x$. Then $frac{Delta y}{Delta x} = 0$ for all choices of $Delta xneq 0$ and this is also the limit as $Delta xto 0$. So the limit in this case is reached (contradicting your claim "can never be reached") and furthermore no value smaller or greater than the limit is ever reached (contradicting your claim "every value smaller or greater to it can be reached").
$endgroup$
– Christoph
Dec 9 '18 at 14:43
$begingroup$
@Christoph: Thanks.... I see.... So constant functions do not fit here. Any other flaws?
$endgroup$
– Joe
Dec 9 '18 at 14:47