Why does h approach 0 and not f(x + h) - f(x) when taking the derivative?
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Why does the blue section approach 0 and not the black one? Also, if h would approach Zero, wouldn't f(x+h)-f(x) too?
derivatives
edited Dec 25 '18 at 21:48
egreg
184k1486205
asked Dec 25 '18 at 21:25
as997as997
1
$endgroup$
add a comment |
$begingroup$
Why does the blue section approach 0 and not the black one? Also, if h would approach Zero, wouldn't f(x+h)-f(x) too?
derivatives
edited Dec 25 '18 at 21:48
egreg
184k1486205
asked Dec 25 '18 at 21:25
as997as997
1
$endgroup$
$begingroup$
Yes it would (for a continuous function). But you define a limit as a variable approaches a value, not as a function of a variable approaches a value.
$endgroup$
– Robert Israel
Dec 25 '18 at 21:29
4
$begingroup$
Both $h$ and $f(x+h)-f(x)$ are approaching zero. But, the ratio of the two does not.
$endgroup$
– D.B.
Dec 25 '18 at 21:29
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@Robert Israel Why, because its easier?
$endgroup$
– as997
Dec 25 '18 at 21:31
$begingroup$
They both tend to $0$ actually.
$endgroup$
– KM101
Dec 25 '18 at 21:33
add a comment |
$begingroup$
Why does the blue section approach 0 and not the black one? Also, if h would approach Zero, wouldn't f(x+h)-f(x) too?
derivatives
edited Dec 25 '18 at 21:48
egreg
184k1486205
asked Dec 25 '18 at 21:25
as997as997
1
$endgroup$
Why does the blue section approach 0 and not the black one? Also, if h would approach Zero, wouldn't f(x+h)-f(x) too?
derivatives
derivatives
edited Dec 25 '18 at 21:48
egreg
184k1486205
asked Dec 25 '18 at 21:25
as997as997
1
edited Dec 25 '18 at 21:48
egreg
184k1486205
asked Dec 25 '18 at 21:25
as997as997
1
edited Dec 25 '18 at 21:48
egreg
184k1486205
edited Dec 25 '18 at 21:48
egreg
184k1486205
edited Dec 25 '18 at 21:48
egreg
184k1486205
184k1486205
asked Dec 25 '18 at 21:25
as997as997
1
asked Dec 25 '18 at 21:25
as997as997
1
asked Dec 25 '18 at 21:25
as997as997
1
1
$begingroup$
Yes it would (for a continuous function). But you define a limit as a variable approaches a value, not as a function of a variable approaches a value.
$endgroup$
– Robert Israel
Dec 25 '18 at 21:29
4
$begingroup$
Both $h$ and $f(x+h)-f(x)$ are approaching zero. But, the ratio of the two does not.
$endgroup$
– D.B.
Dec 25 '18 at 21:29
$begingroup$
@Robert Israel Why, because its easier?
$endgroup$
– as997
Dec 25 '18 at 21:31
$begingroup$
They both tend to $0$ actually.
$endgroup$
– KM101
Dec 25 '18 at 21:33
add a comment |
$begingroup$
Yes it would (for a continuous function). But you define a limit as a variable approaches a value, not as a function of a variable approaches a value.
$endgroup$
– Robert Israel
Dec 25 '18 at 21:29
4
$begingroup$
Both $h$ and $f(x+h)-f(x)$ are approaching zero. But, the ratio of the two does not.
$endgroup$
– D.B.
Dec 25 '18 at 21:29
$begingroup$
@Robert Israel Why, because its easier?
$endgroup$
– as997
Dec 25 '18 at 21:31
$begingroup$
They both tend to $0$ actually.
$endgroup$
– KM101
Dec 25 '18 at 21:33
$begingroup$
Yes it would (for a continuous function). But you define a limit as a variable approaches a value, not as a function of a variable approaches a value.
$endgroup$
– Robert Israel
Dec 25 '18 at 21:29
$begingroup$
Yes it would (for a continuous function). But you define a limit as a variable approaches a value, not as a function of a variable approaches a value.
$endgroup$
– Robert Israel
Dec 25 '18 at 21:29
4
4
$begingroup$
Both $h$ and $f(x+h)-f(x)$ are approaching zero. But, the ratio of the two does not.
$endgroup$
– D.B.
Dec 25 '18 at 21:29
$begingroup$
Both $h$ and $f(x+h)-f(x)$ are approaching zero. But, the ratio of the two does not.
$endgroup$
– D.B.
Dec 25 '18 at 21:29
$begingroup$
@Robert Israel Why, because its easier?
$endgroup$
– as997
Dec 25 '18 at 21:31
$begingroup$
@Robert Israel Why, because its easier?
$endgroup$
– as997
Dec 25 '18 at 21:31
$begingroup$
They both tend to $0$ actually.
$endgroup$
– KM101
Dec 25 '18 at 21:33
$begingroup$
They both tend to $0$ actually.
$endgroup$
– KM101
Dec 25 '18 at 21:33
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The comments have basically answered this, but to make it explicit:
When taking the derivative, we're interested in the limit of the ratio $frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.
If $f$ is continuous at $x$, then both the numerator and the denominator will approach $0$ as $h$ approaches $0$, but the ratio frequently will not.
Explicitly having $f(x+h)-f(x)$ approach $0$ probably wouldn't be useful.
We want the ratio; if we choose a value of $h$ then it's "clear" how to get both the numerator and denominator. If we choose a value of the numerator we might not have a way for calculating the denominator $h$, and in fact a value of $h$ satisfying our choice isn't guaranteed to exist.
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
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1 Answer
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1 Answer
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oldest
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$begingroup$
The comments have basically answered this, but to make it explicit:
When taking the derivative, we're interested in the limit of the ratio $frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.
If $f$ is continuous at $x$, then both the numerator and the denominator will approach $0$ as $h$ approaches $0$, but the ratio frequently will not.
Explicitly having $f(x+h)-f(x)$ approach $0$ probably wouldn't be useful.
We want the ratio; if we choose a value of $h$ then it's "clear" how to get both the numerator and denominator. If we choose a value of the numerator we might not have a way for calculating the denominator $h$, and in fact a value of $h$ satisfying our choice isn't guaranteed to exist.
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
$endgroup$
add a comment |
$begingroup$
The comments have basically answered this, but to make it explicit:
When taking the derivative, we're interested in the limit of the ratio $frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.
If $f$ is continuous at $x$, then both the numerator and the denominator will approach $0$ as $h$ approaches $0$, but the ratio frequently will not.
Explicitly having $f(x+h)-f(x)$ approach $0$ probably wouldn't be useful.
We want the ratio; if we choose a value of $h$ then it's "clear" how to get both the numerator and denominator. If we choose a value of the numerator we might not have a way for calculating the denominator $h$, and in fact a value of $h$ satisfying our choice isn't guaranteed to exist.
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
$endgroup$
add a comment |
$begingroup$
The comments have basically answered this, but to make it explicit:
When taking the derivative, we're interested in the limit of the ratio $frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.
If $f$ is continuous at $x$, then both the numerator and the denominator will approach $0$ as $h$ approaches $0$, but the ratio frequently will not.
Explicitly having $f(x+h)-f(x)$ approach $0$ probably wouldn't be useful.
We want the ratio; if we choose a value of $h$ then it's "clear" how to get both the numerator and denominator. If we choose a value of the numerator we might not have a way for calculating the denominator $h$, and in fact a value of $h$ satisfying our choice isn't guaranteed to exist.
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
$endgroup$
The comments have basically answered this, but to make it explicit:
When taking the derivative, we're interested in the limit of the ratio $frac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.
If $f$ is continuous at $x$, then both the numerator and the denominator will approach $0$ as $h$ approaches $0$, but the ratio frequently will not.
Explicitly having $f(x+h)-f(x)$ approach $0$ probably wouldn't be useful.
We want the ratio; if we choose a value of $h$ then it's "clear" how to get both the numerator and denominator. If we choose a value of the numerator we might not have a way for calculating the denominator $h$, and in fact a value of $h$ satisfying our choice isn't guaranteed to exist.
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
answered Dec 25 '18 at 22:06
ShapeOfMatterShapeOfMatter
1794
1794
add a comment |
add a comment |
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$begingroup$
Yes it would (for a continuous function). But you define a limit as a variable approaches a value, not as a function of a variable approaches a value.
$endgroup$
– Robert Israel
Dec 25 '18 at 21:29
4
$begingroup$
Both $h$ and $f(x+h)-f(x)$ are approaching zero. But, the ratio of the two does not.
$endgroup$
– D.B.
Dec 25 '18 at 21:29
$begingroup$
@Robert Israel Why, because its easier?
$endgroup$
– as997
Dec 25 '18 at 21:31
$begingroup$
They both tend to $0$ actually.
$endgroup$
– KM101
Dec 25 '18 at 21:33