Whats Wrong with this approach?

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I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.



$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$



solving the above as follows:



$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$



$=lim_xto0(frac3x^4−frac3x^4)$



$=lim_xto0(0)$



$=0$




calculus limits




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  • 2




    How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
    – Lord Shark the Unknown
    Aug 2 at 8:13







  • 3




    You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
    – Mauro ALLEGRANZA
    Aug 2 at 8:14










  • Using individual limits on $x^4$ and $fracsinxx$
    – Ashwani Bhat
    Aug 2 at 8:15






  • 3




    You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
    – Kavi Rama Murthy
    Aug 2 at 8:24















up vote
0
down vote

favorite












I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.



$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$



solving the above as follows:



$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$



$=lim_xto0(frac3x^4−frac3x^4)$



$=lim_xto0(0)$



$=0$




calculus limits




share|cite|improve this question















  • 2




    How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
    – Lord Shark the Unknown
    Aug 2 at 8:13







  • 3




    You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
    – Mauro ALLEGRANZA
    Aug 2 at 8:14










  • Using individual limits on $x^4$ and $fracsinxx$
    – Ashwani Bhat
    Aug 2 at 8:15






  • 3




    You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
    – Kavi Rama Murthy
    Aug 2 at 8:24













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.



$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$



solving the above as follows:



$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$



$=lim_xto0(frac3x^4−frac3x^4)$



$=lim_xto0(0)$



$=0$




calculus limits




share|cite|improve this question











I saw this question few moments back in here. The answer to this question is not 0 and is proceeded using Taylor expansion. I want to know where the error is in the below approach.



$lim_xto0 (frac2+cosxx^3sinx−frac3x^4)$



solving the above as follows:



$=lim_xto0(frac2x^3sinx+fraccosxx^3sinx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+fraccosxx^4fracsinxx−frac3x^4)$



$=lim_xto0(frac2x^4fracsinxx+frac1x^4fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4lim_xto0fracsinxx+frac1x^4lim_xto0fractanxx−frac3x^4)$



$=lim_xto0(frac2x^4+frac1x^4−frac3x^4)$



$=lim_xto0(frac3x^4−frac3x^4)$



$=lim_xto0(0)$



$=0$





calculus limits





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share|cite|improve this question




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asked Aug 2 at 8:10









Ashwani Bhat

1




1







  • 2




    How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
    – Lord Shark the Unknown
    Aug 2 at 8:13







  • 3




    You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
    – Mauro ALLEGRANZA
    Aug 2 at 8:14










  • Using individual limits on $x^4$ and $fracsinxx$
    – Ashwani Bhat
    Aug 2 at 8:15






  • 3




    You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
    – Kavi Rama Murthy
    Aug 2 at 8:24













  • 2




    How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
    – Lord Shark the Unknown
    Aug 2 at 8:13







  • 3




    You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
    – Mauro ALLEGRANZA
    Aug 2 at 8:14










  • Using individual limits on $x^4$ and $fracsinxx$
    – Ashwani Bhat
    Aug 2 at 8:15






  • 3




    You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
    – Kavi Rama Murthy
    Aug 2 at 8:24








2




2




How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13





How do you go from $lim_xto0left(frac2x^4fracsin xx+frac1x^4fractan xx−frac3x^4right)$ to the next line?
– Lord Shark the Unknown
Aug 2 at 8:13





3




3




You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14




You cannot "move out" $x^4$ from limit, because the limit is about $x$ itself.
– Mauro ALLEGRANZA
Aug 2 at 8:14












Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15




Using individual limits on $x^4$ and $fracsinxx$
– Ashwani Bhat
Aug 2 at 8:15




3




3




You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24





You will have to have limit for $x^4$ also which makes the first teem infinite. You cannot write $lim frac sin x x=lim frac lim sin x x=lim frac 0 x=0$, right?
– Kavi Rama Murthy
Aug 2 at 8:24
















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