How to solve $lim left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2}$












2












$begingroup$


I can't seem to find a way to solve:



$$lim left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}$$



I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings light to the solution.



Any ideas?






sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: put $n=1/t$ where $t>0$ is small.
    $endgroup$
    – user1892304
    Jan 25 at 11:56










  • $begingroup$
    No dfrac in titles please.
    $endgroup$
    – Did
    Jan 25 at 12:32










  • $begingroup$
    limits at which point? you didn't mention the point.... :(
    $endgroup$
    – Abhas Kumar Sinha
    Jan 25 at 15:25










  • $begingroup$
    @AbhasKumarSinha, when using $n$ we usually assume $n to infty$ so I omitted that part.
    $endgroup$
    – Concept7
    Jan 25 at 17:32
















2












$begingroup$


I can't seem to find a way to solve:



$$lim left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}$$



I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings light to the solution.



Any ideas?






sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: put $n=1/t$ where $t>0$ is small.
    $endgroup$
    – user1892304
    Jan 25 at 11:56










  • $begingroup$
    No dfrac in titles please.
    $endgroup$
    – Did
    Jan 25 at 12:32










  • $begingroup$
    limits at which point? you didn't mention the point.... :(
    $endgroup$
    – Abhas Kumar Sinha
    Jan 25 at 15:25










  • $begingroup$
    @AbhasKumarSinha, when using $n$ we usually assume $n to infty$ so I omitted that part.
    $endgroup$
    – Concept7
    Jan 25 at 17:32














2












2








2





$begingroup$


I can't seem to find a way to solve:



$$lim left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}$$



I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings light to the solution.



Any ideas?






sequences-and-series limits







share|cite|improve this question











$endgroup$




I can't seem to find a way to solve:



$$lim left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}$$



I've tried applying an exponential and logaritmic to take the $n^2$ out of the exponent, I've tried dividing the expression, but I don't get anywhere that brings light to the solution.



Any ideas?






sequences-and-series limits




sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 12:32









Did

248k23225463




248k23225463










asked Jan 25 at 11:53









Concept7Concept7

1468




1468












  • $begingroup$
    Hint: put $n=1/t$ where $t>0$ is small.
    $endgroup$
    – user1892304
    Jan 25 at 11:56










  • $begingroup$
    No dfrac in titles please.
    $endgroup$
    – Did
    Jan 25 at 12:32










  • $begingroup$
    limits at which point? you didn't mention the point.... :(
    $endgroup$
    – Abhas Kumar Sinha
    Jan 25 at 15:25










  • $begingroup$
    @AbhasKumarSinha, when using $n$ we usually assume $n to infty$ so I omitted that part.
    $endgroup$
    – Concept7
    Jan 25 at 17:32


















  • $begingroup$
    Hint: put $n=1/t$ where $t>0$ is small.
    $endgroup$
    – user1892304
    Jan 25 at 11:56










  • $begingroup$
    No dfrac in titles please.
    $endgroup$
    – Did
    Jan 25 at 12:32










  • $begingroup$
    limits at which point? you didn't mention the point.... :(
    $endgroup$
    – Abhas Kumar Sinha
    Jan 25 at 15:25










  • $begingroup$
    @AbhasKumarSinha, when using $n$ we usually assume $n to infty$ so I omitted that part.
    $endgroup$
    – Concept7
    Jan 25 at 17:32
















$begingroup$
Hint: put $n=1/t$ where $t>0$ is small.
$endgroup$
– user1892304
Jan 25 at 11:56




$begingroup$
Hint: put $n=1/t$ where $t>0$ is small.
$endgroup$
– user1892304
Jan 25 at 11:56












$begingroup$
No dfrac in titles please.
$endgroup$
– Did
Jan 25 at 12:32




$begingroup$
No dfrac in titles please.
$endgroup$
– Did
Jan 25 at 12:32












$begingroup$
limits at which point? you didn't mention the point.... :(
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:25




$begingroup$
limits at which point? you didn't mention the point.... :(
$endgroup$
– Abhas Kumar Sinha
Jan 25 at 15:25












$begingroup$
@AbhasKumarSinha, when using $n$ we usually assume $n to infty$ so I omitted that part.
$endgroup$
– Concept7
Jan 25 at 17:32




$begingroup$
@AbhasKumarSinha, when using $n$ we usually assume $n to infty$ so I omitted that part.
$endgroup$
– Concept7
Jan 25 at 17:32










4 Answers
4






active

oldest

votes


















1












$begingroup$

$$
begin{align*}
L &= lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} \
&= lim_{ntoinfty}left(frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}right)^{n^2}tag1 \
&= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{n^2} tag2 \
&= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} tag3 \
&= lim_{ntoinfty}e^{n^2(n-2n^2+4)over(n^3+2n^2)} tag4
end{align*}
$$
Now consider:
$$
lim_{ntoinfty}{n^2(n-2n^2+4)over(n^3+2n^2)} = -infty
$$



Hence your limit is:
$$
e^{-infty} = 0
$$



Description of steps:





  • $(1)$ add and subtract $2n^2$


  • $(2)$ perform division


  • $(3)$ multiply the power by the reciprocal of the fraction inside parentheses


  • $(4)$ use the limit for $(1 + {1over x^n})^{x_n}$ when $x_n to infty$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
    $endgroup$
    – P. Quinton
    Jan 25 at 12:13












  • $begingroup$
    @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
    $endgroup$
    – roman
    Jan 25 at 12:13












  • $begingroup$
    I was actually talking about the step just before that, the fourth
    $endgroup$
    – P. Quinton
    Jan 25 at 12:14










  • $begingroup$
    @P.Quinton well that would be a good candidate for the OP to consider proving it
    $endgroup$
    – roman
    Jan 25 at 12:24



















1












$begingroup$

$$lim_{n rightarrow infty} log f(n) = n^2 logdfrac{n^3+n+4}{n^3+2n^2}= n^2 log (1 - O(frac{1}{n})) rightarrow -infty $$
Hence $$lim_{n rightarrow infty} f(n) = e^{-infty} = 0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I suppose that $nto infty$



    $$lim_{nto infty} left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}=lim_{nto infty} Biggl(left(1+dfrac{-2n^2+n+4}{n^3+2n^2}right)^{dfrac{n^3+2n^2}{-2n^2+n+4}}Biggr)^{frac{-2n^4+n^3+4n^2}{n^3+2n^2}}= e^{-infty}=0$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      We have $$ lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} =
      lim_{ntoinfty}left(frac{1+frac{n+4}{n^3} }{1+frac{2n^2}{n^3}}right)^{n^2}=
      frac{e}{lim_{ntoinfty}e^n}$$



      Hence your limit is: 0






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        $$
        begin{align*}
        L &= lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} \
        &= lim_{ntoinfty}left(frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}right)^{n^2}tag1 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{n^2} tag2 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} tag3 \
        &= lim_{ntoinfty}e^{n^2(n-2n^2+4)over(n^3+2n^2)} tag4
        end{align*}
        $$
        Now consider:
        $$
        lim_{ntoinfty}{n^2(n-2n^2+4)over(n^3+2n^2)} = -infty
        $$



        Hence your limit is:
        $$
        e^{-infty} = 0
        $$



        Description of steps:





        • $(1)$ add and subtract $2n^2$


        • $(2)$ perform division


        • $(3)$ multiply the power by the reciprocal of the fraction inside parentheses


        • $(4)$ use the limit for $(1 + {1over x^n})^{x_n}$ when $x_n to infty$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
          $endgroup$
          – P. Quinton
          Jan 25 at 12:13












        • $begingroup$
          @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
          $endgroup$
          – roman
          Jan 25 at 12:13












        • $begingroup$
          I was actually talking about the step just before that, the fourth
          $endgroup$
          – P. Quinton
          Jan 25 at 12:14










        • $begingroup$
          @P.Quinton well that would be a good candidate for the OP to consider proving it
          $endgroup$
          – roman
          Jan 25 at 12:24
















        1












        $begingroup$

        $$
        begin{align*}
        L &= lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} \
        &= lim_{ntoinfty}left(frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}right)^{n^2}tag1 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{n^2} tag2 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} tag3 \
        &= lim_{ntoinfty}e^{n^2(n-2n^2+4)over(n^3+2n^2)} tag4
        end{align*}
        $$
        Now consider:
        $$
        lim_{ntoinfty}{n^2(n-2n^2+4)over(n^3+2n^2)} = -infty
        $$



        Hence your limit is:
        $$
        e^{-infty} = 0
        $$



        Description of steps:





        • $(1)$ add and subtract $2n^2$


        • $(2)$ perform division


        • $(3)$ multiply the power by the reciprocal of the fraction inside parentheses


        • $(4)$ use the limit for $(1 + {1over x^n})^{x_n}$ when $x_n to infty$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
          $endgroup$
          – P. Quinton
          Jan 25 at 12:13












        • $begingroup$
          @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
          $endgroup$
          – roman
          Jan 25 at 12:13












        • $begingroup$
          I was actually talking about the step just before that, the fourth
          $endgroup$
          – P. Quinton
          Jan 25 at 12:14










        • $begingroup$
          @P.Quinton well that would be a good candidate for the OP to consider proving it
          $endgroup$
          – roman
          Jan 25 at 12:24














        1












        1








        1





        $begingroup$

        $$
        begin{align*}
        L &= lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} \
        &= lim_{ntoinfty}left(frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}right)^{n^2}tag1 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{n^2} tag2 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} tag3 \
        &= lim_{ntoinfty}e^{n^2(n-2n^2+4)over(n^3+2n^2)} tag4
        end{align*}
        $$
        Now consider:
        $$
        lim_{ntoinfty}{n^2(n-2n^2+4)over(n^3+2n^2)} = -infty
        $$



        Hence your limit is:
        $$
        e^{-infty} = 0
        $$



        Description of steps:





        • $(1)$ add and subtract $2n^2$


        • $(2)$ perform division


        • $(3)$ multiply the power by the reciprocal of the fraction inside parentheses


        • $(4)$ use the limit for $(1 + {1over x^n})^{x_n}$ when $x_n to infty$






        share|cite|improve this answer











        $endgroup$



        $$
        begin{align*}
        L &= lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} \
        &= lim_{ntoinfty}left(frac{n^3 +2n^2 - 2n^2+n+4}{n^3+2n^2}right)^{n^2}tag1 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{n^2} tag2 \
        &= lim_{ntoinfty}left(1 + frac{- 2n^2+n+4}{n^3+2n^2}right)^{frac{n^2(n^3+2n^2)(n-2n^2+4)}{(n^3+2n^2)(n-2n^2+4)}} tag3 \
        &= lim_{ntoinfty}e^{n^2(n-2n^2+4)over(n^3+2n^2)} tag4
        end{align*}
        $$
        Now consider:
        $$
        lim_{ntoinfty}{n^2(n-2n^2+4)over(n^3+2n^2)} = -infty
        $$



        Hence your limit is:
        $$
        e^{-infty} = 0
        $$



        Description of steps:





        • $(1)$ add and subtract $2n^2$


        • $(2)$ perform division


        • $(3)$ multiply the power by the reciprocal of the fraction inside parentheses


        • $(4)$ use the limit for $(1 + {1over x^n})^{x_n}$ when $x_n to infty$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 25 at 12:11

























        answered Jan 25 at 12:07









        romanroman

        2,34321224




        2,34321224












        • $begingroup$
          Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
          $endgroup$
          – P. Quinton
          Jan 25 at 12:13












        • $begingroup$
          @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
          $endgroup$
          – roman
          Jan 25 at 12:13












        • $begingroup$
          I was actually talking about the step just before that, the fourth
          $endgroup$
          – P. Quinton
          Jan 25 at 12:14










        • $begingroup$
          @P.Quinton well that would be a good candidate for the OP to consider proving it
          $endgroup$
          – roman
          Jan 25 at 12:24


















        • $begingroup$
          Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
          $endgroup$
          – P. Quinton
          Jan 25 at 12:13












        • $begingroup$
          @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
          $endgroup$
          – roman
          Jan 25 at 12:13












        • $begingroup$
          I was actually talking about the step just before that, the fourth
          $endgroup$
          – P. Quinton
          Jan 25 at 12:14










        • $begingroup$
          @P.Quinton well that would be a good candidate for the OP to consider proving it
          $endgroup$
          – roman
          Jan 25 at 12:24
















        $begingroup$
        Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
        $endgroup$
        – P. Quinton
        Jan 25 at 12:13






        $begingroup$
        Some extra work need to be taken care of in step 4, the limit is $(1+frac{1}{x^n})^{x_n y_n}to e^{y_n}$ when $x_nto infty$, this is not trivial and I'm not sure it is always true.
        $endgroup$
        – P. Quinton
        Jan 25 at 12:13














        $begingroup$
        @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
        $endgroup$
        – roman
        Jan 25 at 12:13






        $begingroup$
        @P.Quinton you may justify this by continuity of $a^x$. Hence $lim a^{x_n} = a^{lim x_n}$
        $endgroup$
        – roman
        Jan 25 at 12:13














        $begingroup$
        I was actually talking about the step just before that, the fourth
        $endgroup$
        – P. Quinton
        Jan 25 at 12:14




        $begingroup$
        I was actually talking about the step just before that, the fourth
        $endgroup$
        – P. Quinton
        Jan 25 at 12:14












        $begingroup$
        @P.Quinton well that would be a good candidate for the OP to consider proving it
        $endgroup$
        – roman
        Jan 25 at 12:24




        $begingroup$
        @P.Quinton well that would be a good candidate for the OP to consider proving it
        $endgroup$
        – roman
        Jan 25 at 12:24











        1












        $begingroup$

        $$lim_{n rightarrow infty} log f(n) = n^2 logdfrac{n^3+n+4}{n^3+2n^2}= n^2 log (1 - O(frac{1}{n})) rightarrow -infty $$
        Hence $$lim_{n rightarrow infty} f(n) = e^{-infty} = 0$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          $$lim_{n rightarrow infty} log f(n) = n^2 logdfrac{n^3+n+4}{n^3+2n^2}= n^2 log (1 - O(frac{1}{n})) rightarrow -infty $$
          Hence $$lim_{n rightarrow infty} f(n) = e^{-infty} = 0$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            $$lim_{n rightarrow infty} log f(n) = n^2 logdfrac{n^3+n+4}{n^3+2n^2}= n^2 log (1 - O(frac{1}{n})) rightarrow -infty $$
            Hence $$lim_{n rightarrow infty} f(n) = e^{-infty} = 0$$






            share|cite|improve this answer









            $endgroup$



            $$lim_{n rightarrow infty} log f(n) = n^2 logdfrac{n^3+n+4}{n^3+2n^2}= n^2 log (1 - O(frac{1}{n})) rightarrow -infty $$
            Hence $$lim_{n rightarrow infty} f(n) = e^{-infty} = 0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 25 at 12:05









            Ahmad BazziAhmad Bazzi

            8,3622824




            8,3622824























                1












                $begingroup$

                I suppose that $nto infty$



                $$lim_{nto infty} left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}=lim_{nto infty} Biggl(left(1+dfrac{-2n^2+n+4}{n^3+2n^2}right)^{dfrac{n^3+2n^2}{-2n^2+n+4}}Biggr)^{frac{-2n^4+n^3+4n^2}{n^3+2n^2}}= e^{-infty}=0$$






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  I suppose that $nto infty$



                  $$lim_{nto infty} left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}=lim_{nto infty} Biggl(left(1+dfrac{-2n^2+n+4}{n^3+2n^2}right)^{dfrac{n^3+2n^2}{-2n^2+n+4}}Biggr)^{frac{-2n^4+n^3+4n^2}{n^3+2n^2}}= e^{-infty}=0$$






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    I suppose that $nto infty$



                    $$lim_{nto infty} left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}=lim_{nto infty} Biggl(left(1+dfrac{-2n^2+n+4}{n^3+2n^2}right)^{dfrac{n^3+2n^2}{-2n^2+n+4}}Biggr)^{frac{-2n^4+n^3+4n^2}{n^3+2n^2}}= e^{-infty}=0$$






                    share|cite|improve this answer









                    $endgroup$



                    I suppose that $nto infty$



                    $$lim_{nto infty} left(dfrac{n^3+n+4}{n^3+2n^2}right)^{n^2}=lim_{nto infty} Biggl(left(1+dfrac{-2n^2+n+4}{n^3+2n^2}right)^{dfrac{n^3+2n^2}{-2n^2+n+4}}Biggr)^{frac{-2n^4+n^3+4n^2}{n^3+2n^2}}= e^{-infty}=0$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 25 at 12:13









                    J.DaneJ.Dane

                    368114




                    368114























                        1












                        $begingroup$

                        We have $$ lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} =
                        lim_{ntoinfty}left(frac{1+frac{n+4}{n^3} }{1+frac{2n^2}{n^3}}right)^{n^2}=
                        frac{e}{lim_{ntoinfty}e^n}$$



                        Hence your limit is: 0






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          We have $$ lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} =
                          lim_{ntoinfty}left(frac{1+frac{n+4}{n^3} }{1+frac{2n^2}{n^3}}right)^{n^2}=
                          frac{e}{lim_{ntoinfty}e^n}$$



                          Hence your limit is: 0






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            We have $$ lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} =
                            lim_{ntoinfty}left(frac{1+frac{n+4}{n^3} }{1+frac{2n^2}{n^3}}right)^{n^2}=
                            frac{e}{lim_{ntoinfty}e^n}$$



                            Hence your limit is: 0






                            share|cite|improve this answer









                            $endgroup$



                            We have $$ lim_{ntoinfty}left(frac{n^3+n+4}{n^3+2n^2}right)^{n^2} =
                            lim_{ntoinfty}left(frac{1+frac{n+4}{n^3} }{1+frac{2n^2}{n^3}}right)^{n^2}=
                            frac{e}{lim_{ntoinfty}e^n}$$



                            Hence your limit is: 0







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 25 at 12:21









                            user62498user62498

                            1,978614




                            1,978614






























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