Mixing
Evaluating $lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$
$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$
calculus limits radicals limits-without-lhopital
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
$endgroup$
|
show 6 more comments
$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$
calculus limits radicals limits-without-lhopital
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
$endgroup$
$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40
$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nəsibli
Jan 14 at 16:44
$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17
$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18
$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nəsibli
Jan 14 at 17:18
|
show 6 more comments
$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$
calculus limits radicals limits-without-lhopital
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
$endgroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
I think,it would be useful to start with simplifying the first term in the numerator as $(sqrt{x+x^3}-x)=x^{{3over2}}$ and leaving the denominator untouched.Then,i have to just work on the remaining term,but I can't figure out a way to not have $ln1=0$
calculus limits radicals limits-without-lhopital
calculus limits radicals limits-without-lhopital
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
edited Jan 14 at 18:56
Michael Rozenberg
103k1891196
103k1891196
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
asked Jan 14 at 16:36
Turan NəsibliTuran Nəsibli
776
776
$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40
$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nəsibli
Jan 14 at 16:44
$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17
$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18
$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nəsibli
Jan 14 at 17:18
|
show 6 more comments
$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40
$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nəsibli
Jan 14 at 16:44
$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17
$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18
$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nəsibli
Jan 14 at 17:18
$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40
$begingroup$
Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
$endgroup$
– coreyman317
Jan 14 at 16:40
$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nəsibli
Jan 14 at 16:44
$begingroup$
As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
$endgroup$
– Turan Nəsibli
Jan 14 at 16:44
$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17
$begingroup$
While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
$endgroup$
– JimB
Jan 14 at 17:17
$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18
$begingroup$
Just because they are asymptotically equivalent does not mean they are simply equal.
$endgroup$
– Simply Beautiful Art
Jan 14 at 17:18
$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nəsibli
Jan 14 at 17:18
$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nəsibli
Jan 14 at 17:18
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
add a comment |
$begingroup$
$$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$
Set $x=1/h^2,hto0^+$
$$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$
$$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$
$2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$
$$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
add a comment |
$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
add a comment |
$begingroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
answered Jan 14 at 18:54
Michael RozenbergMichael Rozenberg
103k1891196
103k1891196
add a comment |
add a comment |
$begingroup$
$$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$
Set $x=1/h^2,hto0^+$
$$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$
$$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$
$2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$
$$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
$$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$
Set $x=1/h^2,hto0^+$
$$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$
$$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$
$2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$
$$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
$$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$
Set $x=1/h^2,hto0^+$
$$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$
$$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$
$2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$
$$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$
Set $x=1/h^2,hto0^+$
$$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$
$$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$
$2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$
$$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
answered Jan 17 at 14:52
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
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Note: $sqrt{x+x^3}-xneq x^{frac{3}{2}}$
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– coreyman317
Jan 14 at 16:40
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As $xto+infty$ $x$ is negligible with respect to $x^3$ and it becomes $sqrt {x^3}=x^{{3over2}}$ and then isn't $x$ negligible with respect to $x^{{3over2}}$?
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– Turan Nəsibli
Jan 14 at 16:44
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While I'm sure you need the steps to get to the answer, it sometimes helps if one knows the answer which in this case is $-3/pi$.
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– JimB
Jan 14 at 17:17
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Just because they are asymptotically equivalent does not mean they are simply equal.
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– Simply Beautiful Art
Jan 14 at 17:18
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@JimB I know the answer as well,thanks anyway.
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– Turan Nəsibli
Jan 14 at 17:18