What's the difference between the two？
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Which of the following two can define the derivative $f'(a)$:
1)$$lim_{n to infty}n left[fleft(a+frac{1}{n}right)-f(a)right],n in mathbb{Z}.$$
2)$$lim_{x to infty}xleft[fleft(a+frac{1}{x}right)-f(a)right],x in mathbb{R}.$$
derivatives
asked Nov 15 at 14:02
mengdie1982
4,272618
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up vote
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Question
Which of the following two can define the derivative $f'(a)$:
1)$$lim_{n to infty}n left[fleft(a+frac{1}{n}right)-f(a)right],n in mathbb{Z}.$$
2)$$lim_{x to infty}xleft[fleft(a+frac{1}{x}right)-f(a)right],x in mathbb{R}.$$
derivatives
asked Nov 15 at 14:02
mengdie1982
4,272618
Here, the fact $limlimits_{x to infty}$ exists implies that $limlimits_{x to +infty}$ and $limlimits_{x to -infty}$ both exist and equal.
– mengdie1982
Nov 15 at 14:10
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative.
– Paramanand Singh
Nov 15 at 14:34
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh
– mengdie1982
Nov 15 at 14:36
Refer to this answer for substitution rule of limits: math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity.
– Paramanand Singh
Nov 15 at 14:39
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values.
– Paramanand Singh
Nov 15 at 14:41
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question
Which of the following two can define the derivative $f'(a)$:
1)$$lim_{n to infty}n left[fleft(a+frac{1}{n}right)-f(a)right],n in mathbb{Z}.$$
2)$$lim_{x to infty}xleft[fleft(a+frac{1}{x}right)-f(a)right],x in mathbb{R}.$$
derivatives
asked Nov 15 at 14:02
mengdie1982
4,272618
Question
Which of the following two can define the derivative $f'(a)$:
1)$$lim_{n to infty}n left[fleft(a+frac{1}{n}right)-f(a)right],n in mathbb{Z}.$$
2)$$lim_{x to infty}xleft[fleft(a+frac{1}{x}right)-f(a)right],x in mathbb{R}.$$
derivatives
derivatives
asked Nov 15 at 14:02
mengdie1982
4,272618
asked Nov 15 at 14:02
mengdie1982
4,272618
asked Nov 15 at 14:02
mengdie1982
4,272618
asked Nov 15 at 14:02
mengdie1982
4,272618
asked Nov 15 at 14:02
mengdie1982
4,272618
4,272618
Here, the fact $limlimits_{x to infty}$ exists implies that $limlimits_{x to +infty}$ and $limlimits_{x to -infty}$ both exist and equal.
– mengdie1982
Nov 15 at 14:10
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative.
– Paramanand Singh
Nov 15 at 14:34
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh
– mengdie1982
Nov 15 at 14:36
Refer to this answer for substitution rule of limits: math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity.
– Paramanand Singh
Nov 15 at 14:39
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values.
– Paramanand Singh
Nov 15 at 14:41
|
show 1 more comment
Here, the fact $limlimits_{x to infty}$ exists implies that $limlimits_{x to +infty}$ and $limlimits_{x to -infty}$ both exist and equal.
– mengdie1982
Nov 15 at 14:10
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative.
– Paramanand Singh
Nov 15 at 14:34
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh
– mengdie1982
Nov 15 at 14:36
Refer to this answer for substitution rule of limits: math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity.
– Paramanand Singh
Nov 15 at 14:39
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values.
– Paramanand Singh
Nov 15 at 14:41
Here, the fact $limlimits_{x to infty}$ exists implies that $limlimits_{x to +infty}$ and $limlimits_{x to -infty}$ both exist and equal.
– mengdie1982
Nov 15 at 14:10
Here, the fact $limlimits_{x to infty}$ exists implies that $limlimits_{x to +infty}$ and $limlimits_{x to -infty}$ both exist and equal.
– mengdie1982
Nov 15 at 14:10
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative.
– Paramanand Singh
Nov 15 at 14:34
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative.
– Paramanand Singh
Nov 15 at 14:34
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh
– mengdie1982
Nov 15 at 14:36
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh
– mengdie1982
Nov 15 at 14:36
Refer to this answer for substitution rule of limits: math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity.
– Paramanand Singh
Nov 15 at 14:39
Refer to this answer for substitution rule of limits: math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity.
– Paramanand Singh
Nov 15 at 14:39
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values.
– Paramanand Singh
Nov 15 at 14:41
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values.
– Paramanand Singh
Nov 15 at 14:41
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
Let $f(x)=
begin{cases}
0 & text{If $x$ is rational.} \
1 & text{If $x$ is irrational}
end{cases}$
Then $displaystyle lim_{substack{n to infty} \ n in mathbb{Z}}n left[fleft(0+frac{1}{n}right)-f(0)right] = 0$
and
$displaystyle lim_{substack{n to infty} \ n in mathbb R setminus mathbb{Q}}n left[fleft(0+frac{1}{n}right)-f(0)right] = infty$.
Hence
$displaystyle lim_{substack{n to infty} \ n in mathbb R}n left[fleft(0+frac{1}{n}right)-f(0)right]$ does not exist.
edited Nov 15 at 14:54
Reinhard Meier
2,777210
answered Nov 15 at 14:25
steven gregory
17.1k22155
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $f(x)=
begin{cases}
0 & text{If $x$ is rational.} \
1 & text{If $x$ is irrational}
end{cases}$
Then $displaystyle lim_{substack{n to infty} \ n in mathbb{Z}}n left[fleft(0+frac{1}{n}right)-f(0)right] = 0$
and
$displaystyle lim_{substack{n to infty} \ n in mathbb R setminus mathbb{Q}}n left[fleft(0+frac{1}{n}right)-f(0)right] = infty$.
Hence
$displaystyle lim_{substack{n to infty} \ n in mathbb R}n left[fleft(0+frac{1}{n}right)-f(0)right]$ does not exist.
edited Nov 15 at 14:54
Reinhard Meier
2,777210
answered Nov 15 at 14:25
steven gregory
17.1k22155
add a comment |
up vote
2
down vote
Let $f(x)=
begin{cases}
0 & text{If $x$ is rational.} \
1 & text{If $x$ is irrational}
end{cases}$
Then $displaystyle lim_{substack{n to infty} \ n in mathbb{Z}}n left[fleft(0+frac{1}{n}right)-f(0)right] = 0$
and
$displaystyle lim_{substack{n to infty} \ n in mathbb R setminus mathbb{Q}}n left[fleft(0+frac{1}{n}right)-f(0)right] = infty$.
Hence
$displaystyle lim_{substack{n to infty} \ n in mathbb R}n left[fleft(0+frac{1}{n}right)-f(0)right]$ does not exist.
edited Nov 15 at 14:54
Reinhard Meier
2,777210
answered Nov 15 at 14:25
steven gregory
17.1k22155
add a comment |
up vote
2
down vote
up vote
2
down vote
Let $f(x)=
begin{cases}
0 & text{If $x$ is rational.} \
1 & text{If $x$ is irrational}
end{cases}$
Then $displaystyle lim_{substack{n to infty} \ n in mathbb{Z}}n left[fleft(0+frac{1}{n}right)-f(0)right] = 0$
and
$displaystyle lim_{substack{n to infty} \ n in mathbb R setminus mathbb{Q}}n left[fleft(0+frac{1}{n}right)-f(0)right] = infty$.
Hence
$displaystyle lim_{substack{n to infty} \ n in mathbb R}n left[fleft(0+frac{1}{n}right)-f(0)right]$ does not exist.
edited Nov 15 at 14:54
Reinhard Meier
2,777210
answered Nov 15 at 14:25
steven gregory
17.1k22155
Let $f(x)=
begin{cases}
0 & text{If $x$ is rational.} \
1 & text{If $x$ is irrational}
end{cases}$
Then $displaystyle lim_{substack{n to infty} \ n in mathbb{Z}}n left[fleft(0+frac{1}{n}right)-f(0)right] = 0$
and
$displaystyle lim_{substack{n to infty} \ n in mathbb R setminus mathbb{Q}}n left[fleft(0+frac{1}{n}right)-f(0)right] = infty$.
Hence
$displaystyle lim_{substack{n to infty} \ n in mathbb R}n left[fleft(0+frac{1}{n}right)-f(0)right]$ does not exist.
edited Nov 15 at 14:54
Reinhard Meier
2,777210
answered Nov 15 at 14:25
steven gregory
17.1k22155
edited Nov 15 at 14:54
Reinhard Meier
2,777210
edited Nov 15 at 14:54
Reinhard Meier
2,777210
edited Nov 15 at 14:54
Reinhard Meier
2,777210
2,777210
answered Nov 15 at 14:25
steven gregory
17.1k22155
answered Nov 15 at 14:25
steven gregory
17.1k22155
answered Nov 15 at 14:25
steven gregory
17.1k22155
17.1k22155
add a comment |
add a comment |
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Here, the fact $limlimits_{x to infty}$ exists implies that $limlimits_{x to +infty}$ and $limlimits_{x to -infty}$ both exist and equal.
– mengdie1982
Nov 15 at 14:10
The second one is the definition of derivative as can be seen by using substitution $x=1/h$. The existence of first limit does not imply the existence of second limit, but existence of second limit implies the existence of first one so the first limit does not match the definition of derivative.
– Paramanand Singh
Nov 15 at 14:34
yes，I agree with you. But when can we make the subsitution?In another word, why we can not substitute $x$ for $1/n$?@ParamanandSingh
– mengdie1982
Nov 15 at 14:36
Refer to this answer for substitution rule of limits: math.stackexchange.com/a/1073047/72031 See the theorem mentioned at the end. It can be modified suitably for the case when variables tend to infinity.
– Paramanand Singh
Nov 15 at 14:39
The $1/n$ can't be substituted by $h$ because $1/n$ takes only a specific set of values namely rationals with numerator $1$ whereas $h$ takes all real values.
– Paramanand Singh
Nov 15 at 14:41