Why The Interval of f(x) is the same in Both One and Two Side Limits?
$begingroup$
My Question might be wired For some But It Will help me to get the idea of One Side Limits.
My Book
Recall The Definition of One Side Limits :
- Right Hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c<x<c+δ quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
- Lift hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c-δ<x<c quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
My Question
It Might Be weird to me That putting an Interval
$$lvert f(x)- llvert<ϵ$$
in Either One of the definitions.
Consider Right hand limit, I'm assume that we need an interval such
$$l<f(x)<l+ϵ quad$$
Because It's Impossible to get any value at $$c<x$$
and The same Thing for Left hand Side , we might put $$l-ϵ<f(x)<l quad$$ instead of $$l-ϵ<f(x)<l+ϵ quad$$ Since it's also impassible for $$c>x$$
What is the point of Use $$lvert f(x)- llvert<ϵ $$ instead of what Just I Wrote of intervals For set the interval of Epsilon?
Thanks in advance
calculus limits
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
$endgroup$
add a comment |
$begingroup$
My Question might be wired For some But It Will help me to get the idea of One Side Limits.
My Book
Recall The Definition of One Side Limits :
- Right Hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c<x<c+δ quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
- Lift hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c-δ<x<c quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
My Question
It Might Be weird to me That putting an Interval
$$lvert f(x)- llvert<ϵ$$
in Either One of the definitions.
Consider Right hand limit, I'm assume that we need an interval such
$$l<f(x)<l+ϵ quad$$
Because It's Impossible to get any value at $$c<x$$
and The same Thing for Left hand Side , we might put $$l-ϵ<f(x)<l quad$$ instead of $$l-ϵ<f(x)<l+ϵ quad$$ Since it's also impassible for $$c>x$$
What is the point of Use $$lvert f(x)- llvert<ϵ $$ instead of what Just I Wrote of intervals For set the interval of Epsilon?
Thanks in advance
calculus limits
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
$endgroup$
1
$begingroup$
What if $f$ is decreasing?
$endgroup$
– Hans Lundmark
Jan 5 at 11:01
$begingroup$
Using $x>c$ does not necessarily imply $f(x) >l$. The right/left refers to values of $x$ with respect to $c$ and not to values of $f(x) $.
$endgroup$
– Paramanand Singh
Jan 5 at 12:52
$begingroup$
@ParamanandSingh- meaning there are cases we will have $$x>c$$ for $$|f(x)-l|<ϵ$$?! Could you give me an example for limit in this case where $$x>c$$ pointing to $$f(x)<l$$
$endgroup$
– Ammar Bamhdi
Jan 6 at 0:36
add a comment |
$begingroup$
My Question might be wired For some But It Will help me to get the idea of One Side Limits.
My Book
Recall The Definition of One Side Limits :
- Right Hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c<x<c+δ quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
- Lift hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c-δ<x<c quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
My Question
It Might Be weird to me That putting an Interval
$$lvert f(x)- llvert<ϵ$$
in Either One of the definitions.
Consider Right hand limit, I'm assume that we need an interval such
$$l<f(x)<l+ϵ quad$$
Because It's Impossible to get any value at $$c<x$$
and The same Thing for Left hand Side , we might put $$l-ϵ<f(x)<l quad$$ instead of $$l-ϵ<f(x)<l+ϵ quad$$ Since it's also impassible for $$c>x$$
What is the point of Use $$lvert f(x)- llvert<ϵ $$ instead of what Just I Wrote of intervals For set the interval of Epsilon?
Thanks in advance
calculus limits
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
$endgroup$
My Question might be wired For some But It Will help me to get the idea of One Side Limits.
My Book
Recall The Definition of One Side Limits :
- Right Hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c<x<c+δ quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
- Lift hand Limit
if for every number ϵ>0 there is a number δ>0 such that
$$c-δ<x<c quadlongrightarrowquad lvert f(x)- llvert<ϵ $$
My Question
It Might Be weird to me That putting an Interval
$$lvert f(x)- llvert<ϵ$$
in Either One of the definitions.
Consider Right hand limit, I'm assume that we need an interval such
$$l<f(x)<l+ϵ quad$$
Because It's Impossible to get any value at $$c<x$$
and The same Thing for Left hand Side , we might put $$l-ϵ<f(x)<l quad$$ instead of $$l-ϵ<f(x)<l+ϵ quad$$ Since it's also impassible for $$c>x$$
What is the point of Use $$lvert f(x)- llvert<ϵ $$ instead of what Just I Wrote of intervals For set the interval of Epsilon?
Thanks in advance
calculus limits
calculus limits
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
asked Jan 5 at 8:11
Ammar BamhdiAmmar Bamhdi
325
325
1
$begingroup$
What if $f$ is decreasing?
$endgroup$
– Hans Lundmark
Jan 5 at 11:01
$begingroup$
Using $x>c$ does not necessarily imply $f(x) >l$. The right/left refers to values of $x$ with respect to $c$ and not to values of $f(x) $.
$endgroup$
– Paramanand Singh
Jan 5 at 12:52
$begingroup$
@ParamanandSingh- meaning there are cases we will have $$x>c$$ for $$|f(x)-l|<ϵ$$?! Could you give me an example for limit in this case where $$x>c$$ pointing to $$f(x)<l$$
$endgroup$
– Ammar Bamhdi
Jan 6 at 0:36
add a comment |
1
$begingroup$
What if $f$ is decreasing?
$endgroup$
– Hans Lundmark
Jan 5 at 11:01
$begingroup$
Using $x>c$ does not necessarily imply $f(x) >l$. The right/left refers to values of $x$ with respect to $c$ and not to values of $f(x) $.
$endgroup$
– Paramanand Singh
Jan 5 at 12:52
$begingroup$
@ParamanandSingh- meaning there are cases we will have $$x>c$$ for $$|f(x)-l|<ϵ$$?! Could you give me an example for limit in this case where $$x>c$$ pointing to $$f(x)<l$$
$endgroup$
– Ammar Bamhdi
Jan 6 at 0:36
1
1
$begingroup$
What if $f$ is decreasing?
$endgroup$
– Hans Lundmark
Jan 5 at 11:01
$begingroup$
What if $f$ is decreasing?
$endgroup$
– Hans Lundmark
Jan 5 at 11:01
$begingroup$
Using $x>c$ does not necessarily imply $f(x) >l$. The right/left refers to values of $x$ with respect to $c$ and not to values of $f(x) $.
$endgroup$
– Paramanand Singh
Jan 5 at 12:52
$begingroup$
Using $x>c$ does not necessarily imply $f(x) >l$. The right/left refers to values of $x$ with respect to $c$ and not to values of $f(x) $.
$endgroup$
– Paramanand Singh
Jan 5 at 12:52
$begingroup$
@ParamanandSingh- meaning there are cases we will have $$x>c$$ for $$|f(x)-l|<ϵ$$?! Could you give me an example for limit in this case where $$x>c$$ pointing to $$f(x)<l$$
$endgroup$
– Ammar Bamhdi
Jan 6 at 0:36
$begingroup$
@ParamanandSingh- meaning there are cases we will have $$x>c$$ for $$|f(x)-l|<ϵ$$?! Could you give me an example for limit in this case where $$x>c$$ pointing to $$f(x)<l$$
$endgroup$
– Ammar Bamhdi
Jan 6 at 0:36
add a comment |
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1
$begingroup$
What if $f$ is decreasing?
$endgroup$
– Hans Lundmark
Jan 5 at 11:01
$begingroup$
Using $x>c$ does not necessarily imply $f(x) >l$. The right/left refers to values of $x$ with respect to $c$ and not to values of $f(x) $.
$endgroup$
– Paramanand Singh
Jan 5 at 12:52
$begingroup$
@ParamanandSingh- meaning there are cases we will have $$x>c$$ for $$|f(x)-l|<ϵ$$?! Could you give me an example for limit in this case where $$x>c$$ pointing to $$f(x)<l$$
$endgroup$
– Ammar Bamhdi
Jan 6 at 0:36