Mixing
functions limits
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In question No.9 why we can not conclude anything about f at x=1 ? I did not really get it so can someone explain it for me ?
calculus limits
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
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In question No.9 why we can not conclude anything about f at x=1 ? I did not really get it so can someone explain it for me ?
calculus limits
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
1
Take the function $f(x)=5$ for all real $xne 1$ and an arbitary value for $x=1$. Whatever its value is for $x=1$, the limit for $xrightarrow 1$ is $5$ because if we approach $1$ from either side, the value approaches $5$. With the additional condition that $f$ is continous at $x=1$, we could include $f(1)=5$.
– Peter
Jul 21 at 17:39
As an aside, for problem 10 just reverse the role of the various numbers in the answers below. The end takeaway is that $f(c)$ can be different than $limlimits_xto cf(x)$. Functions where $f(c)=limlimits_xto cf(x)$ for all $c$ are called continuous functions and play a special role in mathematics.
– JMoravitz
Jul 21 at 17:55
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
In question No.9 why we can not conclude anything about f at x=1 ? I did not really get it so can someone explain it for me ?
calculus limits
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
In question No.9 why we can not conclude anything about f at x=1 ? I did not really get it so can someone explain it for me ?
calculus limits
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
asked Jul 21 at 17:36
Ahmed M. Elsonbaty
474
474
1
Take the function $f(x)=5$ for all real $xne 1$ and an arbitary value for $x=1$. Whatever its value is for $x=1$, the limit for $xrightarrow 1$ is $5$ because if we approach $1$ from either side, the value approaches $5$. With the additional condition that $f$ is continous at $x=1$, we could include $f(1)=5$.
– Peter
Jul 21 at 17:39
As an aside, for problem 10 just reverse the role of the various numbers in the answers below. The end takeaway is that $f(c)$ can be different than $limlimits_xto cf(x)$. Functions where $f(c)=limlimits_xto cf(x)$ for all $c$ are called continuous functions and play a special role in mathematics.
– JMoravitz
Jul 21 at 17:55
add a comment |Â
1
Take the function $f(x)=5$ for all real $xne 1$ and an arbitary value for $x=1$. Whatever its value is for $x=1$, the limit for $xrightarrow 1$ is $5$ because if we approach $1$ from either side, the value approaches $5$. With the additional condition that $f$ is continous at $x=1$, we could include $f(1)=5$.
– Peter
Jul 21 at 17:39
As an aside, for problem 10 just reverse the role of the various numbers in the answers below. The end takeaway is that $f(c)$ can be different than $limlimits_xto cf(x)$. Functions where $f(c)=limlimits_xto cf(x)$ for all $c$ are called continuous functions and play a special role in mathematics.
– JMoravitz
Jul 21 at 17:55
1
1
Take the function $f(x)=5$ for all real $xne 1$ and an arbitary value for $x=1$. Whatever its value is for $x=1$, the limit for $xrightarrow 1$ is $5$ because if we approach $1$ from either side, the value approaches $5$. With the additional condition that $f$ is continous at $x=1$, we could include $f(1)=5$.
– Peter
Jul 21 at 17:39
Take the function $f(x)=5$ for all real $xne 1$ and an arbitary value for $x=1$. Whatever its value is for $x=1$, the limit for $xrightarrow 1$ is $5$ because if we approach $1$ from either side, the value approaches $5$. With the additional condition that $f$ is continous at $x=1$, we could include $f(1)=5$.
– Peter
Jul 21 at 17:39
As an aside, for problem 10 just reverse the role of the various numbers in the answers below. The end takeaway is that $f(c)$ can be different than $limlimits_xto cf(x)$. Functions where $f(c)=limlimits_xto cf(x)$ for all $c$ are called continuous functions and play a special role in mathematics.
– JMoravitz
Jul 21 at 17:55
As an aside, for problem 10 just reverse the role of the various numbers in the answers below. The end takeaway is that $f(c)$ can be different than $limlimits_xto cf(x)$. Functions where $f(c)=limlimits_xto cf(x)$ for all $c$ are called continuous functions and play a special role in mathematics.
– JMoravitz
Jul 21 at 17:55
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
In the definition we have $$forall epsilon>0,exists delta, 0<|x-x_0|<deltato |f(x)-L|<epsilon$$for some $L$ where nowhere restricts the value of the function at the exact point. For example consider the following function$$f(x)=begincasesdfrac5x-5x-1&xne 1\a&x=1endcases$$for different values of $a$. In fact if any function satisfies $$lim_xto af(x)=f(a)$$the function is referred to as $textcontinuous at x=a$
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
add a comment |Â
up vote
1
down vote
The definition of limit at $x_0$ is independent of the value of the function at $x_0$. Check the definition!
answered Jul 21 at 17:45
A. Pongrácz
2,263221
add a comment |Â
up vote
1
down vote
Consider the function
$$f(x)~=~begincases
5x~text, for xin mathbbR/1 \
42~text, for x=1 \ endcases $$
while the limit for sure goes for $x=1$ to $5$ the functions is defined in another way at this point. You can construct functions like this with ease and there is the problem.
answered Jul 21 at 17:49
mrtaurho
700219
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In the definition we have $$forall epsilon>0,exists delta, 0<|x-x_0|<deltato |f(x)-L|<epsilon$$for some $L$ where nowhere restricts the value of the function at the exact point. For example consider the following function$$f(x)=begincasesdfrac5x-5x-1&xne 1\a&x=1endcases$$for different values of $a$. In fact if any function satisfies $$lim_xto af(x)=f(a)$$the function is referred to as $textcontinuous at x=a$
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
add a comment |Â
up vote
3
down vote
accepted
In the definition we have $$forall epsilon>0,exists delta, 0<|x-x_0|<deltato |f(x)-L|<epsilon$$for some $L$ where nowhere restricts the value of the function at the exact point. For example consider the following function$$f(x)=begincasesdfrac5x-5x-1&xne 1\a&x=1endcases$$for different values of $a$. In fact if any function satisfies $$lim_xto af(x)=f(a)$$the function is referred to as $textcontinuous at x=a$
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In the definition we have $$forall epsilon>0,exists delta, 0<|x-x_0|<deltato |f(x)-L|<epsilon$$for some $L$ where nowhere restricts the value of the function at the exact point. For example consider the following function$$f(x)=begincasesdfrac5x-5x-1&xne 1\a&x=1endcases$$for different values of $a$. In fact if any function satisfies $$lim_xto af(x)=f(a)$$the function is referred to as $textcontinuous at x=a$
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
In the definition we have $$forall epsilon>0,exists delta, 0<|x-x_0|<deltato |f(x)-L|<epsilon$$for some $L$ where nowhere restricts the value of the function at the exact point. For example consider the following function$$f(x)=begincasesdfrac5x-5x-1&xne 1\a&x=1endcases$$for different values of $a$. In fact if any function satisfies $$lim_xto af(x)=f(a)$$the function is referred to as $textcontinuous at x=a$
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
edited Jul 21 at 18:41
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
answered Jul 21 at 17:50
Mostafa Ayaz
8,5773630
8,5773630
add a comment |Â
add a comment |Â
up vote
1
down vote
The definition of limit at $x_0$ is independent of the value of the function at $x_0$. Check the definition!
answered Jul 21 at 17:45
A. Pongrácz
2,263221
add a comment |Â
up vote
1
down vote
The definition of limit at $x_0$ is independent of the value of the function at $x_0$. Check the definition!
answered Jul 21 at 17:45
A. Pongrácz
2,263221
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The definition of limit at $x_0$ is independent of the value of the function at $x_0$. Check the definition!
answered Jul 21 at 17:45
A. Pongrácz
2,263221
The definition of limit at $x_0$ is independent of the value of the function at $x_0$. Check the definition!
answered Jul 21 at 17:45
A. Pongrácz
2,263221
answered Jul 21 at 17:45
A. Pongrácz
2,263221
answered Jul 21 at 17:45
A. Pongrácz
2,263221
answered Jul 21 at 17:45
A. Pongrácz
2,263221
2,263221
add a comment |Â
add a comment |Â
up vote
1
down vote
Consider the function
$$f(x)~=~begincases
5x~text, for xin mathbbR/1 \
42~text, for x=1 \ endcases $$
while the limit for sure goes for $x=1$ to $5$ the functions is defined in another way at this point. You can construct functions like this with ease and there is the problem.
answered Jul 21 at 17:49
mrtaurho
700219
add a comment |Â
up vote
1
down vote
Consider the function
$$f(x)~=~begincases
5x~text, for xin mathbbR/1 \
42~text, for x=1 \ endcases $$
while the limit for sure goes for $x=1$ to $5$ the functions is defined in another way at this point. You can construct functions like this with ease and there is the problem.
answered Jul 21 at 17:49
mrtaurho
700219
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Consider the function
$$f(x)~=~begincases
5x~text, for xin mathbbR/1 \
42~text, for x=1 \ endcases $$
while the limit for sure goes for $x=1$ to $5$ the functions is defined in another way at this point. You can construct functions like this with ease and there is the problem.
answered Jul 21 at 17:49
mrtaurho
700219
Consider the function
$$f(x)~=~begincases
5x~text, for xin mathbbR/1 \
42~text, for x=1 \ endcases $$
while the limit for sure goes for $x=1$ to $5$ the functions is defined in another way at this point. You can construct functions like this with ease and there is the problem.
answered Jul 21 at 17:49
mrtaurho
700219
answered Jul 21 at 17:49
mrtaurho
700219
answered Jul 21 at 17:49
mrtaurho
700219
answered Jul 21 at 17:49
mrtaurho
700219
700219
add a comment |Â
add a comment |Â
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1
Take the function $f(x)=5$ for all real $xne 1$ and an arbitary value for $x=1$. Whatever its value is for $x=1$, the limit for $xrightarrow 1$ is $5$ because if we approach $1$ from either side, the value approaches $5$. With the additional condition that $f$ is continous at $x=1$, we could include $f(1)=5$.
– Peter
Jul 21 at 17:39
As an aside, for problem 10 just reverse the role of the various numbers in the answers below. The end takeaway is that $f(c)$ can be different than $limlimits_xto cf(x)$. Functions where $f(c)=limlimits_xto cf(x)$ for all $c$ are called continuous functions and play a special role in mathematics.
– JMoravitz
Jul 21 at 17:55