Mixing
Find $lim_{x to -1} 1/(sqrt{|x|-{-x}})$ where ${}$ denotes the fractional part.
$begingroup$
Find $lim_{x to -1} 1/(sqrt{|x|-{-x}})$ where ${}$ denotes the fractional part.
My attempt -
$$lim_{x to -1} frac{1}{sqrt{|x| -{x}+1}}\
lim_{x to -1} frac{1}{sqrt{|x|+1 -x +[x]}}$$
Now,
$$lim_{x to -1+} frac{1}{sqrt{1+1+1 -1}} = frac{1}{sqrt{2}}\
lim_{x to -1-} frac{1}{sqrt{1+1+1-2}} = 1$$
So, limit doesn't exist. But answer is given, that limit exists and is equal to $1$. Where did I go wrong$?$
calculus limits fractional-part
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
$endgroup$
add a comment |
$begingroup$
Find $lim_{x to -1} 1/(sqrt{|x|-{-x}})$ where ${}$ denotes the fractional part.
My attempt -
$$lim_{x to -1} frac{1}{sqrt{|x| -{x}+1}}\
lim_{x to -1} frac{1}{sqrt{|x|+1 -x +[x]}}$$
Now,
$$lim_{x to -1+} frac{1}{sqrt{1+1+1 -1}} = frac{1}{sqrt{2}}\
lim_{x to -1-} frac{1}{sqrt{1+1+1-2}} = 1$$
So, limit doesn't exist. But answer is given, that limit exists and is equal to $1$. Where did I go wrong$?$
calculus limits fractional-part
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
$endgroup$
$begingroup$
How is $|x|-{-x} = |x|-{x}+1$? I don't think so. If $x=frac{1}{3}$ then the LHS is $frac{2}{3}$ while the RHS is $1$.
$endgroup$
– Yanko
Jan 15 at 13:47
$begingroup$
If $xnotinBbb Z,{-x}=1-{x}$. So $|x|-{-x}=|x|-1+{x}$
$endgroup$
– Shubham Johri
Jan 15 at 14:13
add a comment |
$begingroup$
Find $lim_{x to -1} 1/(sqrt{|x|-{-x}})$ where ${}$ denotes the fractional part.
My attempt -
$$lim_{x to -1} frac{1}{sqrt{|x| -{x}+1}}\
lim_{x to -1} frac{1}{sqrt{|x|+1 -x +[x]}}$$
Now,
$$lim_{x to -1+} frac{1}{sqrt{1+1+1 -1}} = frac{1}{sqrt{2}}\
lim_{x to -1-} frac{1}{sqrt{1+1+1-2}} = 1$$
So, limit doesn't exist. But answer is given, that limit exists and is equal to $1$. Where did I go wrong$?$
calculus limits fractional-part
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
$endgroup$
Find $lim_{x to -1} 1/(sqrt{|x|-{-x}})$ where ${}$ denotes the fractional part.
My attempt -
$$lim_{x to -1} frac{1}{sqrt{|x| -{x}+1}}\
lim_{x to -1} frac{1}{sqrt{|x|+1 -x +[x]}}$$
Now,
$$lim_{x to -1+} frac{1}{sqrt{1+1+1 -1}} = frac{1}{sqrt{2}}\
lim_{x to -1-} frac{1}{sqrt{1+1+1-2}} = 1$$
So, limit doesn't exist. But answer is given, that limit exists and is equal to $1$. Where did I go wrong$?$
calculus limits fractional-part
calculus limits fractional-part
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
edited Jan 15 at 15:08
Martin Sleziak
44.7k10118272
44.7k10118272
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
asked Jan 15 at 13:42
MathsaddictMathsaddict
3619
3619
$begingroup$
How is $|x|-{-x} = |x|-{x}+1$? I don't think so. If $x=frac{1}{3}$ then the LHS is $frac{2}{3}$ while the RHS is $1$.
$endgroup$
– Yanko
Jan 15 at 13:47
$begingroup$
If $xnotinBbb Z,{-x}=1-{x}$. So $|x|-{-x}=|x|-1+{x}$
$endgroup$
– Shubham Johri
Jan 15 at 14:13
add a comment |
$begingroup$
How is $|x|-{-x} = |x|-{x}+1$? I don't think so. If $x=frac{1}{3}$ then the LHS is $frac{2}{3}$ while the RHS is $1$.
$endgroup$
– Yanko
Jan 15 at 13:47
$begingroup$
If $xnotinBbb Z,{-x}=1-{x}$. So $|x|-{-x}=|x|-1+{x}$
$endgroup$
– Shubham Johri
Jan 15 at 14:13
$begingroup$
How is $|x|-{-x} = |x|-{x}+1$? I don't think so. If $x=frac{1}{3}$ then the LHS is $frac{2}{3}$ while the RHS is $1$.
$endgroup$
– Yanko
Jan 15 at 13:47
$begingroup$
How is $|x|-{-x} = |x|-{x}+1$? I don't think so. If $x=frac{1}{3}$ then the LHS is $frac{2}{3}$ while the RHS is $1$.
$endgroup$
– Yanko
Jan 15 at 13:47
$begingroup$
If $xnotinBbb Z,{-x}=1-{x}$. So $|x|-{-x}=|x|-1+{x}$
$endgroup$
– Shubham Johri
Jan 15 at 14:13
$begingroup$
If $xnotinBbb Z,{-x}=1-{x}$. So $|x|-{-x}=|x|-1+{x}$
$endgroup$
– Shubham Johri
Jan 15 at 14:13
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In a small vicinity around $x=-1,|x|=-x$. Take $m=-x$, so that the limit transforms to $$lim_{mto1}frac 1{sqrt{m-{m}}}=lim_{mto1}frac 1{sqrt{lfloor mrfloor}}$$Now note that $[0,1)$ does not belong to the domain of $dfrac1{sqrt{lfloor mrfloor}}$, so only one sided-approach, $mto1^+$, is possible. We don't require $mmapstodfrac1{sqrt{lfloor mrfloor}}$ to be defined in the entire vicinity of $1$ in much the same way we claim $lim_{xto0}sqrt x=0$, even though $xmapstosqrt x$ is not defined for $x<0$. Thus, the answer is$$lim_{mto1}frac 1{sqrt{lfloor mrfloor}}=lim_{hto0^+}frac 1{sqrt{lfloor1+hrfloor}}=1$$.
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
$endgroup$
add a comment |
$begingroup$
In the negatives, $|x|=-x$ so that $-x-{-x}=lfloor-xrfloor$. So the function values on the left and on the right are
$$frac1{sqrt1},color{red}{frac1{sqrt0}}$$
and the limit is $1$ (as a limit is computed inside the domain).
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
add a comment |
$begingroup$
If $xin(-2,-1]$, then $|x|-{-x}=-x-(-1-x)=1$.
If $xin(-1,0]$, then $|x|-{-x}=-x-(-x)=0$.
So the left-sided limit exists and equals $1/sqrt 1=1$, but the right-sided limit does not exist $-$ the expression is not even defined on $(-1,0]$.
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
$endgroup$
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In a small vicinity around $x=-1,|x|=-x$. Take $m=-x$, so that the limit transforms to $$lim_{mto1}frac 1{sqrt{m-{m}}}=lim_{mto1}frac 1{sqrt{lfloor mrfloor}}$$Now note that $[0,1)$ does not belong to the domain of $dfrac1{sqrt{lfloor mrfloor}}$, so only one sided-approach, $mto1^+$, is possible. We don't require $mmapstodfrac1{sqrt{lfloor mrfloor}}$ to be defined in the entire vicinity of $1$ in much the same way we claim $lim_{xto0}sqrt x=0$, even though $xmapstosqrt x$ is not defined for $x<0$. Thus, the answer is$$lim_{mto1}frac 1{sqrt{lfloor mrfloor}}=lim_{hto0^+}frac 1{sqrt{lfloor1+hrfloor}}=1$$.
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
$endgroup$
add a comment |
$begingroup$
In a small vicinity around $x=-1,|x|=-x$. Take $m=-x$, so that the limit transforms to $$lim_{mto1}frac 1{sqrt{m-{m}}}=lim_{mto1}frac 1{sqrt{lfloor mrfloor}}$$Now note that $[0,1)$ does not belong to the domain of $dfrac1{sqrt{lfloor mrfloor}}$, so only one sided-approach, $mto1^+$, is possible. We don't require $mmapstodfrac1{sqrt{lfloor mrfloor}}$ to be defined in the entire vicinity of $1$ in much the same way we claim $lim_{xto0}sqrt x=0$, even though $xmapstosqrt x$ is not defined for $x<0$. Thus, the answer is$$lim_{mto1}frac 1{sqrt{lfloor mrfloor}}=lim_{hto0^+}frac 1{sqrt{lfloor1+hrfloor}}=1$$.
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
$endgroup$
add a comment |
$begingroup$
In a small vicinity around $x=-1,|x|=-x$. Take $m=-x$, so that the limit transforms to $$lim_{mto1}frac 1{sqrt{m-{m}}}=lim_{mto1}frac 1{sqrt{lfloor mrfloor}}$$Now note that $[0,1)$ does not belong to the domain of $dfrac1{sqrt{lfloor mrfloor}}$, so only one sided-approach, $mto1^+$, is possible. We don't require $mmapstodfrac1{sqrt{lfloor mrfloor}}$ to be defined in the entire vicinity of $1$ in much the same way we claim $lim_{xto0}sqrt x=0$, even though $xmapstosqrt x$ is not defined for $x<0$. Thus, the answer is$$lim_{mto1}frac 1{sqrt{lfloor mrfloor}}=lim_{hto0^+}frac 1{sqrt{lfloor1+hrfloor}}=1$$.
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
$endgroup$
In a small vicinity around $x=-1,|x|=-x$. Take $m=-x$, so that the limit transforms to $$lim_{mto1}frac 1{sqrt{m-{m}}}=lim_{mto1}frac 1{sqrt{lfloor mrfloor}}$$Now note that $[0,1)$ does not belong to the domain of $dfrac1{sqrt{lfloor mrfloor}}$, so only one sided-approach, $mto1^+$, is possible. We don't require $mmapstodfrac1{sqrt{lfloor mrfloor}}$ to be defined in the entire vicinity of $1$ in much the same way we claim $lim_{xto0}sqrt x=0$, even though $xmapstosqrt x$ is not defined for $x<0$. Thus, the answer is$$lim_{mto1}frac 1{sqrt{lfloor mrfloor}}=lim_{hto0^+}frac 1{sqrt{lfloor1+hrfloor}}=1$$.
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
edited Jan 15 at 14:11
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
answered Jan 15 at 14:05
Shubham JohriShubham Johri
5,177717
5,177717
add a comment |
add a comment |
$begingroup$
In the negatives, $|x|=-x$ so that $-x-{-x}=lfloor-xrfloor$. So the function values on the left and on the right are
$$frac1{sqrt1},color{red}{frac1{sqrt0}}$$
and the limit is $1$ (as a limit is computed inside the domain).
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
add a comment |
$begingroup$
In the negatives, $|x|=-x$ so that $-x-{-x}=lfloor-xrfloor$. So the function values on the left and on the right are
$$frac1{sqrt1},color{red}{frac1{sqrt0}}$$
and the limit is $1$ (as a limit is computed inside the domain).
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
$endgroup$
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
add a comment |
$begingroup$
In the negatives, $|x|=-x$ so that $-x-{-x}=lfloor-xrfloor$. So the function values on the left and on the right are
$$frac1{sqrt1},color{red}{frac1{sqrt0}}$$
and the limit is $1$ (as a limit is computed inside the domain).
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
$endgroup$
In the negatives, $|x|=-x$ so that $-x-{-x}=lfloor-xrfloor$. So the function values on the left and on the right are
$$frac1{sqrt1},color{red}{frac1{sqrt0}}$$
and the limit is $1$ (as a limit is computed inside the domain).
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
edited Jan 15 at 14:21
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
answered Jan 15 at 14:15
Yves DaoustYves Daoust
128k675227
128k675227
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
add a comment |
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
That strikes me as a rather unconventional interpetation! I would not say, for example, that $lim_{xrightarrow 0}sqrt x = 0$.
$endgroup$
– TonyK
Jan 15 at 14:34
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
@TonyK: this is the standard definition of a limit: en.wikipedia.org/wiki/…. Double check your references.
$endgroup$
– Yves Daoust
Jan 15 at 19:52
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
Well, perhaps you are right. But if this were set as an exam question, then I think the examinees would be legitimately annoyed at having to second-guess the examiner's intentions.
$endgroup$
– TonyK
Jan 15 at 20:23
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
$begingroup$
@TonyK: you mean those that skipped school ? Also look at Shubham Johri's answer.
$endgroup$
– Yves Daoust
Jan 15 at 20:45
add a comment |
$begingroup$
If $xin(-2,-1]$, then $|x|-{-x}=-x-(-1-x)=1$.
If $xin(-1,0]$, then $|x|-{-x}=-x-(-x)=0$.
So the left-sided limit exists and equals $1/sqrt 1=1$, but the right-sided limit does not exist $-$ the expression is not even defined on $(-1,0]$.
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
$endgroup$
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
add a comment |
$begingroup$
If $xin(-2,-1]$, then $|x|-{-x}=-x-(-1-x)=1$.
If $xin(-1,0]$, then $|x|-{-x}=-x-(-x)=0$.
So the left-sided limit exists and equals $1/sqrt 1=1$, but the right-sided limit does not exist $-$ the expression is not even defined on $(-1,0]$.
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
$endgroup$
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
add a comment |
$begingroup$
If $xin(-2,-1]$, then $|x|-{-x}=-x-(-1-x)=1$.
If $xin(-1,0]$, then $|x|-{-x}=-x-(-x)=0$.
So the left-sided limit exists and equals $1/sqrt 1=1$, but the right-sided limit does not exist $-$ the expression is not even defined on $(-1,0]$.
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
$endgroup$
If $xin(-2,-1]$, then $|x|-{-x}=-x-(-1-x)=1$.
If $xin(-1,0]$, then $|x|-{-x}=-x-(-x)=0$.
So the left-sided limit exists and equals $1/sqrt 1=1$, but the right-sided limit does not exist $-$ the expression is not even defined on $(-1,0]$.
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
answered Jan 15 at 13:58
TonyKTonyK
42.6k355134
42.6k355134
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
add a comment |
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
$begingroup$
What about the two-sided limit ? ;-)
$endgroup$
– Yves Daoust
Jan 15 at 14:19
add a comment |
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$begingroup$
How is $|x|-{-x} = |x|-{x}+1$? I don't think so. If $x=frac{1}{3}$ then the LHS is $frac{2}{3}$ while the RHS is $1$.
$endgroup$
– Yanko
Jan 15 at 13:47
$begingroup$
If $xnotinBbb Z,{-x}=1-{x}$. So $|x|-{-x}=|x|-1+{x}$
$endgroup$
– Shubham Johri
Jan 15 at 14:13