equivalent definition of limit of a function?
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function and $limlimits_{x to x_0} f(x) = L$. Then I believe the following definiton is equivalent to the definiton of the limit of a function $f(x)$:
for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $0 lt |x-x_0| lt delta$, one has $0lt |f(x) - L| leq epsilon$.
I think its equivalent because of $0 lt |f(x)-L|$ i.e. the existence of the left side of the inequality doesn't matter for the definition of the limit of a function.
Am I right?
limits definition
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
add a comment |
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function and $limlimits_{x to x_0} f(x) = L$. Then I believe the following definiton is equivalent to the definiton of the limit of a function $f(x)$:
for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $0 lt |x-x_0| lt delta$, one has $0lt |f(x) - L| leq epsilon$.
I think its equivalent because of $0 lt |f(x)-L|$ i.e. the existence of the left side of the inequality doesn't matter for the definition of the limit of a function.
Am I right?
limits definition
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
add a comment |
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function and $limlimits_{x to x_0} f(x) = L$. Then I believe the following definiton is equivalent to the definiton of the limit of a function $f(x)$:
for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $0 lt |x-x_0| lt delta$, one has $0lt |f(x) - L| leq epsilon$.
I think its equivalent because of $0 lt |f(x)-L|$ i.e. the existence of the left side of the inequality doesn't matter for the definition of the limit of a function.
Am I right?
limits definition
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
$endgroup$
Let $f:(0,1) to mathbb{R}$ be a given function and $limlimits_{x to x_0} f(x) = L$. Then I believe the following definiton is equivalent to the definiton of the limit of a function $f(x)$:
for any $epsilon gt 0$, there exists $delta gt 0$ such that for all $x in (0,1)$ and $0 lt |x-x_0| lt delta$, one has $0lt |f(x) - L| leq epsilon$.
I think its equivalent because of $0 lt |f(x)-L|$ i.e. the existence of the left side of the inequality doesn't matter for the definition of the limit of a function.
Am I right?
limits definition
limits definition
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
asked Dec 16 '18 at 21:44
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1619
1619
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
This does not work because of the condition $0 < |f(x) - L|$.
Consider a constant function $f(x) = 0$. According to your definition, the limit (with $L=0$) does not exist anywhere since for any $x$, $|f(x) - L| = |0 - 0| = 0 not > 0$.
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
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$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
add a comment |
$begingroup$
That's not equivalent since we can have $f(x)=L$ also for $x neq x_0$ as for example for a constant function, therefore we should state
$$0le |f(x) - L| leq epsilon$$
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
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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
$begingroup$
This does not work because of the condition $0 < |f(x) - L|$.
Consider a constant function $f(x) = 0$. According to your definition, the limit (with $L=0$) does not exist anywhere since for any $x$, $|f(x) - L| = |0 - 0| = 0 not > 0$.
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
$endgroup$
$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
add a comment |
$begingroup$
This does not work because of the condition $0 < |f(x) - L|$.
Consider a constant function $f(x) = 0$. According to your definition, the limit (with $L=0$) does not exist anywhere since for any $x$, $|f(x) - L| = |0 - 0| = 0 not > 0$.
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
$endgroup$
$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
add a comment |
$begingroup$
This does not work because of the condition $0 < |f(x) - L|$.
Consider a constant function $f(x) = 0$. According to your definition, the limit (with $L=0$) does not exist anywhere since for any $x$, $|f(x) - L| = |0 - 0| = 0 not > 0$.
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
$endgroup$
This does not work because of the condition $0 < |f(x) - L|$.
Consider a constant function $f(x) = 0$. According to your definition, the limit (with $L=0$) does not exist anywhere since for any $x$, $|f(x) - L| = |0 - 0| = 0 not > 0$.
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
answered Dec 16 '18 at 21:46
parsiadparsiad
18.3k32453
18.3k32453
$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
add a comment |
$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
that was my first guess, thanks for clarification .
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 16 '18 at 21:49
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
$begingroup$
No problem. Feel free to accept if I definitively answered your question.
$endgroup$
– parsiad
Dec 16 '18 at 21:50
add a comment |
$begingroup$
That's not equivalent since we can have $f(x)=L$ also for $x neq x_0$ as for example for a constant function, therefore we should state
$$0le |f(x) - L| leq epsilon$$
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
That's not equivalent since we can have $f(x)=L$ also for $x neq x_0$ as for example for a constant function, therefore we should state
$$0le |f(x) - L| leq epsilon$$
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
$endgroup$
add a comment |
$begingroup$
That's not equivalent since we can have $f(x)=L$ also for $x neq x_0$ as for example for a constant function, therefore we should state
$$0le |f(x) - L| leq epsilon$$
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
$endgroup$
That's not equivalent since we can have $f(x)=L$ also for $x neq x_0$ as for example for a constant function, therefore we should state
$$0le |f(x) - L| leq epsilon$$
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
answered Dec 16 '18 at 21:50
gimusigimusi
93k84594
93k84594
add a comment |
add a comment |
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