Help find the mistake in this problem of finding limit (using L'Hospital)

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Evaluate $$lim_x to 0 left(frac1x^2-cot^2xright).$$




Attempt
beginalign*
&lim_x to 0 left(frac1x^2-cot^2xright)\
= &lim_x to 0 left(frac1x-cotxright)left(frac1x+cotxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright)left(fracsinx-xcosxxsinxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright) times lim_x to 0left(fracsinx-xcosxxsinxright).
endalign*



Both the terms are in $frac00$ form. So applying L'Hospital on both the limits we have,



$$= lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxsinxxcosx+sinxright).$$



The second term is in $frac00$ form. So applying L'Hospital on the second limit we have,
beginalign*
= &lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxcosx+sinx2cosx-xsinxright)\
=& lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times left(fracxcosx+sinx2cosx-xsinxright)\
=& 1
endalign*



The correct answer is $dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?






limits







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  • The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified.
    – xbh
    3 hours ago














up vote
2
down vote

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Evaluate $$lim_x to 0 left(frac1x^2-cot^2xright).$$




Attempt
beginalign*
&lim_x to 0 left(frac1x^2-cot^2xright)\
= &lim_x to 0 left(frac1x-cotxright)left(frac1x+cotxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright)left(fracsinx-xcosxxsinxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright) times lim_x to 0left(fracsinx-xcosxxsinxright).
endalign*



Both the terms are in $frac00$ form. So applying L'Hospital on both the limits we have,



$$= lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxsinxxcosx+sinxright).$$



The second term is in $frac00$ form. So applying L'Hospital on the second limit we have,
beginalign*
= &lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxcosx+sinx2cosx-xsinxright)\
=& lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times left(fracxcosx+sinx2cosx-xsinxright)\
=& 1
endalign*



The correct answer is $dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?






limits







share|cite|improve this question























  • The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified.
    – xbh
    3 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Evaluate $$lim_x to 0 left(frac1x^2-cot^2xright).$$




Attempt
beginalign*
&lim_x to 0 left(frac1x^2-cot^2xright)\
= &lim_x to 0 left(frac1x-cotxright)left(frac1x+cotxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright)left(fracsinx-xcosxxsinxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright) times lim_x to 0left(fracsinx-xcosxxsinxright).
endalign*



Both the terms are in $frac00$ form. So applying L'Hospital on both the limits we have,



$$= lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxsinxxcosx+sinxright).$$



The second term is in $frac00$ form. So applying L'Hospital on the second limit we have,
beginalign*
= &lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxcosx+sinx2cosx-xsinxright)\
=& lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times left(fracxcosx+sinx2cosx-xsinxright)\
=& 1
endalign*



The correct answer is $dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?






limits







share|cite|improve this question
















Evaluate $$lim_x to 0 left(frac1x^2-cot^2xright).$$




Attempt
beginalign*
&lim_x to 0 left(frac1x^2-cot^2xright)\
= &lim_x to 0 left(frac1x-cotxright)left(frac1x+cotxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright)left(fracsinx-xcosxxsinxright)\
= &lim_x to 0 left(fracsinx+xcosxxsinxright) times lim_x to 0left(fracsinx-xcosxxsinxright).
endalign*



Both the terms are in $frac00$ form. So applying L'Hospital on both the limits we have,



$$= lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxsinxxcosx+sinxright).$$



The second term is in $frac00$ form. So applying L'Hospital on the second limit we have,
beginalign*
= &lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times lim_x to 0left(fracxcosx+sinx2cosx-xsinxright)\
=& lim_x to 0 left(frac2cosx-xsinxxcosx+sinxright) times left(fracxcosx+sinx2cosx-xsinxright)\
=& 1
endalign*



The correct answer is $dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?






limits




limits






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edited 2 hours ago









Tianlalu

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asked 3 hours ago









yathish

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  • The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified.
    – xbh
    3 hours ago
















  • The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified.
    – xbh
    3 hours ago















The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified.
– xbh
3 hours ago




The = on the last 2nd line. Generally speaking, you could do that if both the limit exists. In your case, that is not fully justified.
– xbh
3 hours ago










4 Answers
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Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,



$= lim_x to 0 (frac2cosx-xsinxxcosx+sinx) times lim_x to 0(fracxsinxxcosx+sinx)$




Note that the limit $lim_x to 0 (frac2cosx-xsinxxcosx+sinx)$ does not exist.



In fact, even before that $lim_x to 0 (fracsinx+xcosxxsinx)$ does not exist as well.



This is the graph of $fracsinx+xcosxxsinx$.



enter image description here






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    beginalignlim_xto0frac1x+cot x&=lim_xto0fracsin x+xcos xxsin x\&=lim_xto0frac2cos x-xsin xsin x+xcos x\&=infty.endalignBesides,$$lim_xto0frac1x-cot x=0.$$






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      $$lim_xto0left(dfrac1x^2-dfrac1tan^2xright)=lim_xto0dfractan x-xx^3cdotlim_xto0dfractan x+ xxcdotleft(lim_xto0dfrac xtan xright)^2$$



      Now the limit for the last two are too simple for L'hospital



      For $lim_xto0dfractan x-xx^3$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion






      share|cite|improve this answer
















      • 1




        I don't think this fully answers OP's question.
        – é«˜ç”°èˆª
        3 hours ago

















      up vote
      0
      down vote













      Similar work:
      $$beginalignlim_x to 0 left(frac1x^2-cot^2xright)&=lim_x to 0 left(frac1x^2-frac1-sin^2xsin^2xright)=\
      &=lim_x to 0left(frac1x^2-frac1sin^2x+1right)=\
      &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xright)+1=\
      &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xcdot fracsin^2xx^2right)+1=\
      &=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1.
      endalign$$
      According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
      $$beginalignlim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1=&\
      underbracelim_x to 0left(fracsin x-xx^3right)_-frac16cdot underbracelim_x to 0left(fracsin x+xxright)_=2 +1stackrelLR=&\
      lim_x to 0left(fraccos x-13x^2right)cdot lim_x to 0left(fraccos x+11right) +1stackrelLR=&\
      lim_x to 0left(frac-sin x6xright)cdot 2 +1=&\
      -frac13+1=&frac23.endalign$$
      However you can not express:
      $$-frac13=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)=\
      underbracelim_x to 0left(fracsin x-xx^colorred0right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue4right)_=infty textOR\
      underbracelim_x to 0left(fracsin x-xx^colorred1right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue3right)_=infty textOR\
      underbracelim_x to 0left(fracsin x-xx^colorred2right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue2right)_=infty textOR\
      underbracelim_x to 0left(fracsin x-xx^colorred4right)_=inftycdot underbracelim_x to 0left(fracsin x+xx^colorblue0right)_=0.$$
      because all are the indeterminate form of $0cdot infty$.






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        4 Answers
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        4 Answers
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        Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,



        $= lim_x to 0 (frac2cosx-xsinxxcosx+sinx) times lim_x to 0(fracxsinxxcosx+sinx)$




        Note that the limit $lim_x to 0 (frac2cosx-xsinxxcosx+sinx)$ does not exist.



        In fact, even before that $lim_x to 0 (fracsinx+xcosxxsinx)$ does not exist as well.



        This is the graph of $fracsinx+xcosxxsinx$.



        enter image description here






        share|cite|improve this answer
























          up vote
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          Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,



          $= lim_x to 0 (frac2cosx-xsinxxcosx+sinx) times lim_x to 0(fracxsinxxcosx+sinx)$




          Note that the limit $lim_x to 0 (frac2cosx-xsinxxcosx+sinx)$ does not exist.



          In fact, even before that $lim_x to 0 (fracsinx+xcosxxsinx)$ does not exist as well.



          This is the graph of $fracsinx+xcosxxsinx$.



          enter image description here






          share|cite|improve this answer






















            up vote
            3
            down vote










            up vote
            3
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            Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,



            $= lim_x to 0 (frac2cosx-xsinxxcosx+sinx) times lim_x to 0(fracxsinxxcosx+sinx)$




            Note that the limit $lim_x to 0 (frac2cosx-xsinxxcosx+sinx)$ does not exist.



            In fact, even before that $lim_x to 0 (fracsinx+xcosxxsinx)$ does not exist as well.



            This is the graph of $fracsinx+xcosxxsinx$.



            enter image description here






            share|cite|improve this answer













            Both the terms are in 0/0 form. So applying L'Hospital on both the limits we have,



            $= lim_x to 0 (frac2cosx-xsinxxcosx+sinx) times lim_x to 0(fracxsinxxcosx+sinx)$




            Note that the limit $lim_x to 0 (frac2cosx-xsinxxcosx+sinx)$ does not exist.



            In fact, even before that $lim_x to 0 (fracsinx+xcosxxsinx)$ does not exist as well.



            This is the graph of $fracsinx+xcosxxsinx$.



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 2 hours ago









            Siong Thye Goh

            88.6k1460111




            88.6k1460111




















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                beginalignlim_xto0frac1x+cot x&=lim_xto0fracsin x+xcos xxsin x\&=lim_xto0frac2cos x-xsin xsin x+xcos x\&=infty.endalignBesides,$$lim_xto0frac1x-cot x=0.$$






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                  up vote
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                  beginalignlim_xto0frac1x+cot x&=lim_xto0fracsin x+xcos xxsin x\&=lim_xto0frac2cos x-xsin xsin x+xcos x\&=infty.endalignBesides,$$lim_xto0frac1x-cot x=0.$$






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                    beginalignlim_xto0frac1x+cot x&=lim_xto0fracsin x+xcos xxsin x\&=lim_xto0frac2cos x-xsin xsin x+xcos x\&=infty.endalignBesides,$$lim_xto0frac1x-cot x=0.$$






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                    beginalignlim_xto0frac1x+cot x&=lim_xto0fracsin x+xcos xxsin x\&=lim_xto0frac2cos x-xsin xsin x+xcos x\&=infty.endalignBesides,$$lim_xto0frac1x-cot x=0.$$







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                    answered 3 hours ago









                    José Carlos Santos

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                        $$lim_xto0left(dfrac1x^2-dfrac1tan^2xright)=lim_xto0dfractan x-xx^3cdotlim_xto0dfractan x+ xxcdotleft(lim_xto0dfrac xtan xright)^2$$



                        Now the limit for the last two are too simple for L'hospital



                        For $lim_xto0dfractan x-xx^3$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion






                        share|cite|improve this answer
















                        • 1




                          I don't think this fully answers OP's question.
                          – é«˜ç”°èˆª
                          3 hours ago














                        up vote
                        0
                        down vote













                        $$lim_xto0left(dfrac1x^2-dfrac1tan^2xright)=lim_xto0dfractan x-xx^3cdotlim_xto0dfractan x+ xxcdotleft(lim_xto0dfrac xtan xright)^2$$



                        Now the limit for the last two are too simple for L'hospital



                        For $lim_xto0dfractan x-xx^3$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion






                        share|cite|improve this answer
















                        • 1




                          I don't think this fully answers OP's question.
                          – é«˜ç”°èˆª
                          3 hours ago












                        up vote
                        0
                        down vote










                        up vote
                        0
                        down vote









                        $$lim_xto0left(dfrac1x^2-dfrac1tan^2xright)=lim_xto0dfractan x-xx^3cdotlim_xto0dfractan x+ xxcdotleft(lim_xto0dfrac xtan xright)^2$$



                        Now the limit for the last two are too simple for L'hospital



                        For $lim_xto0dfractan x-xx^3$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion






                        share|cite|improve this answer












                        $$lim_xto0left(dfrac1x^2-dfrac1tan^2xright)=lim_xto0dfractan x-xx^3cdotlim_xto0dfractan x+ xxcdotleft(lim_xto0dfrac xtan xright)^2$$



                        Now the limit for the last two are too simple for L'hospital



                        For $lim_xto0dfractan x-xx^3$ either use L'hospital or Are all limits solvable without L'Hôpital Rule or Series Expansion







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 3 hours ago









                        lab bhattacharjee

                        218k14153268




                        218k14153268







                        • 1




                          I don't think this fully answers OP's question.
                          – é«˜ç”°èˆª
                          3 hours ago












                        • 1




                          I don't think this fully answers OP's question.
                          – é«˜ç”°èˆª
                          3 hours ago







                        1




                        1




                        I don't think this fully answers OP's question.
                        – é«˜ç”°èˆª
                        3 hours ago




                        I don't think this fully answers OP's question.
                        – é«˜ç”°èˆª
                        3 hours ago










                        up vote
                        0
                        down vote













                        Similar work:
                        $$beginalignlim_x to 0 left(frac1x^2-cot^2xright)&=lim_x to 0 left(frac1x^2-frac1-sin^2xsin^2xright)=\
                        &=lim_x to 0left(frac1x^2-frac1sin^2x+1right)=\
                        &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xright)+1=\
                        &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xcdot fracsin^2xx^2right)+1=\
                        &=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1.
                        endalign$$
                        According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
                        $$beginalignlim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1=&\
                        underbracelim_x to 0left(fracsin x-xx^3right)_-frac16cdot underbracelim_x to 0left(fracsin x+xxright)_=2 +1stackrelLR=&\
                        lim_x to 0left(fraccos x-13x^2right)cdot lim_x to 0left(fraccos x+11right) +1stackrelLR=&\
                        lim_x to 0left(frac-sin x6xright)cdot 2 +1=&\
                        -frac13+1=&frac23.endalign$$
                        However you can not express:
                        $$-frac13=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)=\
                        underbracelim_x to 0left(fracsin x-xx^colorred0right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue4right)_=infty textOR\
                        underbracelim_x to 0left(fracsin x-xx^colorred1right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue3right)_=infty textOR\
                        underbracelim_x to 0left(fracsin x-xx^colorred2right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue2right)_=infty textOR\
                        underbracelim_x to 0left(fracsin x-xx^colorred4right)_=inftycdot underbracelim_x to 0left(fracsin x+xx^colorblue0right)_=0.$$
                        because all are the indeterminate form of $0cdot infty$.






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                          Similar work:
                          $$beginalignlim_x to 0 left(frac1x^2-cot^2xright)&=lim_x to 0 left(frac1x^2-frac1-sin^2xsin^2xright)=\
                          &=lim_x to 0left(frac1x^2-frac1sin^2x+1right)=\
                          &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xright)+1=\
                          &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xcdot fracsin^2xx^2right)+1=\
                          &=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1.
                          endalign$$
                          According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
                          $$beginalignlim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1=&\
                          underbracelim_x to 0left(fracsin x-xx^3right)_-frac16cdot underbracelim_x to 0left(fracsin x+xxright)_=2 +1stackrelLR=&\
                          lim_x to 0left(fraccos x-13x^2right)cdot lim_x to 0left(fraccos x+11right) +1stackrelLR=&\
                          lim_x to 0left(frac-sin x6xright)cdot 2 +1=&\
                          -frac13+1=&frac23.endalign$$
                          However you can not express:
                          $$-frac13=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)=\
                          underbracelim_x to 0left(fracsin x-xx^colorred0right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue4right)_=infty textOR\
                          underbracelim_x to 0left(fracsin x-xx^colorred1right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue3right)_=infty textOR\
                          underbracelim_x to 0left(fracsin x-xx^colorred2right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue2right)_=infty textOR\
                          underbracelim_x to 0left(fracsin x-xx^colorred4right)_=inftycdot underbracelim_x to 0left(fracsin x+xx^colorblue0right)_=0.$$
                          because all are the indeterminate form of $0cdot infty$.






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                            up vote
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                            Similar work:
                            $$beginalignlim_x to 0 left(frac1x^2-cot^2xright)&=lim_x to 0 left(frac1x^2-frac1-sin^2xsin^2xright)=\
                            &=lim_x to 0left(frac1x^2-frac1sin^2x+1right)=\
                            &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xright)+1=\
                            &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xcdot fracsin^2xx^2right)+1=\
                            &=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1.
                            endalign$$
                            According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
                            $$beginalignlim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1=&\
                            underbracelim_x to 0left(fracsin x-xx^3right)_-frac16cdot underbracelim_x to 0left(fracsin x+xxright)_=2 +1stackrelLR=&\
                            lim_x to 0left(fraccos x-13x^2right)cdot lim_x to 0left(fraccos x+11right) +1stackrelLR=&\
                            lim_x to 0left(frac-sin x6xright)cdot 2 +1=&\
                            -frac13+1=&frac23.endalign$$
                            However you can not express:
                            $$-frac13=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)=\
                            underbracelim_x to 0left(fracsin x-xx^colorred0right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue4right)_=infty textOR\
                            underbracelim_x to 0left(fracsin x-xx^colorred1right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue3right)_=infty textOR\
                            underbracelim_x to 0left(fracsin x-xx^colorred2right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue2right)_=infty textOR\
                            underbracelim_x to 0left(fracsin x-xx^colorred4right)_=inftycdot underbracelim_x to 0left(fracsin x+xx^colorblue0right)_=0.$$
                            because all are the indeterminate form of $0cdot infty$.






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                            Similar work:
                            $$beginalignlim_x to 0 left(frac1x^2-cot^2xright)&=lim_x to 0 left(frac1x^2-frac1-sin^2xsin^2xright)=\
                            &=lim_x to 0left(frac1x^2-frac1sin^2x+1right)=\
                            &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xright)+1=\
                            &=lim_x to 0left(frac(sin x-x)(sin x+x)x^2sin^2xcdot fracsin^2xx^2right)+1=\
                            &=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1.
                            endalign$$
                            According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
                            $$beginalignlim_x to 0left(frac(sin x-x)(sin x+x)x^4right)+1=&\
                            underbracelim_x to 0left(fracsin x-xx^3right)_-frac16cdot underbracelim_x to 0left(fracsin x+xxright)_=2 +1stackrelLR=&\
                            lim_x to 0left(fraccos x-13x^2right)cdot lim_x to 0left(fraccos x+11right) +1stackrelLR=&\
                            lim_x to 0left(frac-sin x6xright)cdot 2 +1=&\
                            -frac13+1=&frac23.endalign$$
                            However you can not express:
                            $$-frac13=lim_x to 0left(frac(sin x-x)(sin x+x)x^4right)=\
                            underbracelim_x to 0left(fracsin x-xx^colorred0right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue4right)_=infty textOR\
                            underbracelim_x to 0left(fracsin x-xx^colorred1right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue3right)_=infty textOR\
                            underbracelim_x to 0left(fracsin x-xx^colorred2right)_=0cdot underbracelim_x to 0left(fracsin x+xx^colorblue2right)_=infty textOR\
                            underbracelim_x to 0left(fracsin x-xx^colorred4right)_=inftycdot underbracelim_x to 0left(fracsin x+xx^colorblue0right)_=0.$$
                            because all are the indeterminate form of $0cdot infty$.







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                            answered 2 hours ago









                            farruhota

                            16.6k2735




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