explain why the following definition is not equivalent to the definition of the limit of function f
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$ there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - L| < varepsilon$ .
I don't see the difference between this definition and the actual definition of the limit of a function. Could someone help me out with this?
Please and thank you.
limits definition
edited Dec 16 '18 at 22:28
Community♦
1
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
$endgroup$
add a comment |
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$ there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - L| < varepsilon$ .
I don't see the difference between this definition and the actual definition of the limit of a function. Could someone help me out with this?
Please and thank you.
limits definition
edited Dec 16 '18 at 22:28
Community♦
1
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
$endgroup$
2
$begingroup$
You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤delta$. If, say, your function is $f(x)=1$ for $xneq frac 12$ and $fleft( frac 12right) =0$ then $lim_{xto frac 12}f(x)=1$ with the usual definition, but not with your definition.
$endgroup$
– lulu
Dec 11 '18 at 0:56
$begingroup$
Wait so do you mean to say that it’s suppose to be $0 lt |x- x_0| lt delta$ And not less than or equal to delta?
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 11 '18 at 18:40
1
$begingroup$
That's not a meaningful difference. Think it through...you can always modify $delta$ according to which definition you favor. But the $>0$ requirement has real meaning.
$endgroup$
– lulu
Dec 11 '18 at 19:05
add a comment |
$begingroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$ there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - L| < varepsilon$ .
I don't see the difference between this definition and the actual definition of the limit of a function. Could someone help me out with this?
Please and thank you.
limits definition
edited Dec 16 '18 at 22:28
Community♦
1
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
$endgroup$
Let $f:(0,1) to mathbb{R}$ be a given function. Explain how the following definition is not equivalent to the definition of the limit
$limlimits_{x to x_0} f(x) = L$
of $f$ at $x_0 in [0,1]$ .
For any $epsilon gt 0$ there exists $delta gt 0$ such that for all $x in (0,1)$ and $|x-x_0| leq delta$, one has $|f(x) - L| < varepsilon$ .
I don't see the difference between this definition and the actual definition of the limit of a function. Could someone help me out with this?
Please and thank you.
limits definition
limits definition
edited Dec 16 '18 at 22:28
Community♦
1
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
edited Dec 16 '18 at 22:28
Community♦
1
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
edited Dec 16 '18 at 22:28
Community♦
1
edited Dec 16 '18 at 22:28
Community♦
1
edited Dec 16 '18 at 22:28
Community♦
1
1
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
asked Dec 11 '18 at 0:53
ISuckAtMathPleaseHELPMEISuckAtMathPleaseHELPME
1549
1549
2
$begingroup$
You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤delta$. If, say, your function is $f(x)=1$ for $xneq frac 12$ and $fleft( frac 12right) =0$ then $lim_{xto frac 12}f(x)=1$ with the usual definition, but not with your definition.
$endgroup$
– lulu
Dec 11 '18 at 0:56
$begingroup$
Wait so do you mean to say that it’s suppose to be $0 lt |x- x_0| lt delta$ And not less than or equal to delta?
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 11 '18 at 18:40
1
$begingroup$
That's not a meaningful difference. Think it through...you can always modify $delta$ according to which definition you favor. But the $>0$ requirement has real meaning.
$endgroup$
– lulu
Dec 11 '18 at 19:05
add a comment |
2
$begingroup$
You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤delta$. If, say, your function is $f(x)=1$ for $xneq frac 12$ and $fleft( frac 12right) =0$ then $lim_{xto frac 12}f(x)=1$ with the usual definition, but not with your definition.
$endgroup$
– lulu
Dec 11 '18 at 0:56
$begingroup$
Wait so do you mean to say that it’s suppose to be $0 lt |x- x_0| lt delta$ And not less than or equal to delta?
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 11 '18 at 18:40
1
$begingroup$
That's not a meaningful difference. Think it through...you can always modify $delta$ according to which definition you favor. But the $>0$ requirement has real meaning.
$endgroup$
– lulu
Dec 11 '18 at 19:05
2
2
$begingroup$
You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤delta$. If, say, your function is $f(x)=1$ for $xneq frac 12$ and $fleft( frac 12right) =0$ then $lim_{xto frac 12}f(x)=1$ with the usual definition, but not with your definition.
$endgroup$
– lulu
Dec 11 '18 at 0:56
$begingroup$
You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤delta$. If, say, your function is $f(x)=1$ for $xneq frac 12$ and $fleft( frac 12right) =0$ then $lim_{xto frac 12}f(x)=1$ with the usual definition, but not with your definition.
$endgroup$
– lulu
Dec 11 '18 at 0:56
$begingroup$
Wait so do you mean to say that it’s suppose to be $0 lt |x- x_0| lt delta$ And not less than or equal to delta?
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 11 '18 at 18:40
$begingroup$
Wait so do you mean to say that it’s suppose to be $0 lt |x- x_0| lt delta$ And not less than or equal to delta?
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 11 '18 at 18:40
1
1
$begingroup$
That's not a meaningful difference. Think it through...you can always modify $delta$ according to which definition you favor. But the $>0$ requirement has real meaning.
$endgroup$
– lulu
Dec 11 '18 at 19:05
$begingroup$
That's not a meaningful difference. Think it through...you can always modify $delta$ according to which definition you favor. But the $>0$ requirement has real meaning.
$endgroup$
– lulu
Dec 11 '18 at 19:05
add a comment |
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$begingroup$
You have to exclude $x=x_0$, that's why the standard definition has $0<|x-x_0|≤delta$. If, say, your function is $f(x)=1$ for $xneq frac 12$ and $fleft( frac 12right) =0$ then $lim_{xto frac 12}f(x)=1$ with the usual definition, but not with your definition.
$endgroup$
– lulu
Dec 11 '18 at 0:56
$begingroup$
Wait so do you mean to say that it’s suppose to be $0 lt |x- x_0| lt delta$ And not less than or equal to delta?
$endgroup$
– ISuckAtMathPleaseHELPME
Dec 11 '18 at 18:40
1
$begingroup$
That's not a meaningful difference. Think it through...you can always modify $delta$ according to which definition you favor. But the $>0$ requirement has real meaning.
$endgroup$
– lulu
Dec 11 '18 at 19:05