Evaluating $lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$












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$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
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







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  • $begingroup$
    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 Nasibli
    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 Nasibli
    Jan 14 at 17:18
















0












$begingroup$


$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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 Nasibli
    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 Nasibli
    Jan 14 at 17:18














0












0








0





$begingroup$


$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
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







share|cite|improve this question











$endgroup$




$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}$$
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






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edited Jan 14 at 18:56









Michael Rozenberg

109k1896200




109k1896200










asked Jan 14 at 16:36









Turan NasibliTuran Nasibli

846




846












  • $begingroup$
    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 Nasibli
    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 Nasibli
    Jan 14 at 17:18


















  • $begingroup$
    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 Nasibli
    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 Nasibli
    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 Nasibli
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 Nasibli
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 Nasibli
Jan 14 at 17:18




$begingroup$
@JimB I know the answer as well,thanks anyway.
$endgroup$
– Turan Nasibli
Jan 14 at 17:18










2 Answers
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$begingroup$

$$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
$$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
$$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
$$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






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    $begingroup$

    $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



    Set $x=1/h^2,hto0^+$



    $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



    $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



    $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



    $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






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      2 Answers
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      2 Answers
      2






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      active

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      active

      oldest

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      2












      $begingroup$

      $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
      $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
      $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
      $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






      share|cite|improve this answer









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        2












        $begingroup$

        $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
        $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
        $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
        $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
          $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
          $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
          $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$






          share|cite|improve this answer









          $endgroup$



          $$lim_{xto+infty}frac{(sqrt{x+x^3}-x)lnfrac{sqrt{4x+1}}{2sqrt x +3}}{xarctan x}=lim_{xto+infty}frac{x^3lnleft(frac{sqrt{4x+1}}{2sqrt x +3}-1+1right)}{(sqrt{x+x^3}-x)xarctan x}=$$
          $$=lim_{xto+infty}frac{x^2left(frac{sqrt{4x+1}}{2sqrt x +3}-1right)}{(sqrt{x+x^3}-x)arctan x}=lim_{xto+infty}frac{x^2(4x+1-4x-12sqrt{x}-9)}{(2sqrt x +3)(sqrt{4x+1}+2sqrt x +3)(sqrt{x+x^3}-x)arctan x}=$$
          $$=lim_{xto+infty}frac{-12-frac{8}{sqrt{x}}}{left(2 +frac{3}{sqrt{x}}right)left(sqrt{4+frac{1}{x}}+2 +frac{3}{sqrt{x}}right)left(sqrt{frac{1}{x^2}+1}-frac{1}{sqrt{x}}right)arctan x}=$$
          $$=frac{-12}{2cdot4cdot1cdotfrac{pi}{2}}=-frac{3}{pi}.$$







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          answered Jan 14 at 18:54









          Michael RozenbergMichael Rozenberg

          109k1896200




          109k1896200























              0












              $begingroup$

              $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



              Set $x=1/h^2,hto0^+$



              $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



              $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



              $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



              $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






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              $endgroup$


















                0












                $begingroup$

                $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



                Set $x=1/h^2,hto0^+$



                $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



                $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



                $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



                $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






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                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



                  Set $x=1/h^2,hto0^+$



                  $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



                  $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



                  $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



                  $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$






                  share|cite|improve this answer









                  $endgroup$



                  $$I=lim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}{xarctan x}=dfrac2pilim_{xto+infty}frac{(sqrt{x+x^3}-x)ln(frac{sqrt{4x+1}}{2sqrt x +3})}x$$



                  Set $x=1/h^2,hto0^+$



                  $$sqrt{x+x^3}-x=sqrt{dfrac1{h^2}+dfrac1{h^6}}-dfrac{1}{h^2}=dfrac{sqrt{1+h^4}-h}{h^3}text{ and }dfrac{sqrt{4x+1}}{2sqrt{x}+3}= dfrac{sqrt{4+h^2}}{2+3h} $$



                  $$dfrac{pi I}2=lim_{hto0^+}dfrac{h^2(sqrt{1+h^4}-h)}{h^3}lnleft(dfrac{sqrt{4+h^2}}{2+3h}right)$$



                  $2lnleft(dfrac{sqrt{4+h^2}}{2+3 h}right)=lndfrac{4+h^2}{(2+3h)^2}=lnleft(1-dfrac{12h}{(2+3h)^2}right)$



                  $$pi I=-lim_{hto0^+}(sqrt{1+h^4}-h)cdotlim_{hto0^+}left(lim_{hto0^+}dfrac{lnleft(1-dfrac{12h}{(2+3h)^2}right)}{-dfrac{12h}{(2+3h)^2}}right)cdotlim_{hto0^+}dfrac{dfrac{12h}{(2+3h)^2}}h=-1cdot1cdotdfrac{12}4$$







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                  answered Jan 17 at 14:52









                  lab bhattacharjeelab bhattacharjee

                  227k15158278




                  227k15158278






























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                      Questions related to Moebius Transform of Characteristic Function of the Primes

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                      2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
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                      List of scandals in India

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                      List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
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                      Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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                      0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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