Evaluating $limlimits_{x to infty}frac{4^{x+2}+3^x}{4^{x-2}}$












6












$begingroup$



$$lim_{x→∞}frac{4^{x+2}+3^x}{4^{x-2}}.$$




I have solved it like below:
$$lim_{x→∞}left(frac{4^{x+2}}{4^{x-2}}+frac{3^x}{4^{x-2}}right)=lim_{x→∞}left(4^4+frac{3^x}{4^x}·4^2right).$$
Since, as $x → ∞$, $3^x → ∞$, $dfrac{3^x}{4^x} → 0$, the limit is equal to $4^4=256$.



Have I solved it correctly?



This was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.






calculus limits







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$endgroup$








  • 2




    $begingroup$
    $lim_{x to infty} left( 4^4+frac{3^x}{4^x} times 4^2 right)=lim_{x to infty}4^4 +lim_{x to infty}left(dfrac{3^x}{4^x}right)4^2$? While its true you might want to explain for rigor.
    $endgroup$
    – Yadati Kiran
    Dec 22 '18 at 4:26












  • $begingroup$
    It's unclear to me what purpose it serves to write $3^xtoinfty$. It makes the next expression, $frac{3^x}{4^x} to 0$, look wrong even though it actually is correct.
    $endgroup$
    – David K
    Jan 8 at 14:28


















6












$begingroup$



$$lim_{x→∞}frac{4^{x+2}+3^x}{4^{x-2}}.$$




I have solved it like below:
$$lim_{x→∞}left(frac{4^{x+2}}{4^{x-2}}+frac{3^x}{4^{x-2}}right)=lim_{x→∞}left(4^4+frac{3^x}{4^x}·4^2right).$$
Since, as $x → ∞$, $3^x → ∞$, $dfrac{3^x}{4^x} → 0$, the limit is equal to $4^4=256$.



Have I solved it correctly?



This was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.






calculus limits







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $lim_{x to infty} left( 4^4+frac{3^x}{4^x} times 4^2 right)=lim_{x to infty}4^4 +lim_{x to infty}left(dfrac{3^x}{4^x}right)4^2$? While its true you might want to explain for rigor.
    $endgroup$
    – Yadati Kiran
    Dec 22 '18 at 4:26












  • $begingroup$
    It's unclear to me what purpose it serves to write $3^xtoinfty$. It makes the next expression, $frac{3^x}{4^x} to 0$, look wrong even though it actually is correct.
    $endgroup$
    – David K
    Jan 8 at 14:28
















6












6








6


1



$begingroup$



$$lim_{x→∞}frac{4^{x+2}+3^x}{4^{x-2}}.$$




I have solved it like below:
$$lim_{x→∞}left(frac{4^{x+2}}{4^{x-2}}+frac{3^x}{4^{x-2}}right)=lim_{x→∞}left(4^4+frac{3^x}{4^x}·4^2right).$$
Since, as $x → ∞$, $3^x → ∞$, $dfrac{3^x}{4^x} → 0$, the limit is equal to $4^4=256$.



Have I solved it correctly?



This was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.






calculus limits







share|cite|improve this question











$endgroup$





$$lim_{x→∞}frac{4^{x+2}+3^x}{4^{x-2}}.$$




I have solved it like below:
$$lim_{x→∞}left(frac{4^{x+2}}{4^{x-2}}+frac{3^x}{4^{x-2}}right)=lim_{x→∞}left(4^4+frac{3^x}{4^x}·4^2right).$$
Since, as $x → ∞$, $3^x → ∞$, $dfrac{3^x}{4^x} → 0$, the limit is equal to $4^4=256$.



Have I solved it correctly?



This was a practice test question and the given solution was wrong. So, I solved it and I am preparing alone, no friend to discuss, so I posted it here.






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 22 '18 at 7:00









Saad

19.7k92352




19.7k92352










asked Dec 22 '18 at 4:14









user3767495user3767495

3888




3888








  • 2




    $begingroup$
    $lim_{x to infty} left( 4^4+frac{3^x}{4^x} times 4^2 right)=lim_{x to infty}4^4 +lim_{x to infty}left(dfrac{3^x}{4^x}right)4^2$? While its true you might want to explain for rigor.
    $endgroup$
    – Yadati Kiran
    Dec 22 '18 at 4:26












  • $begingroup$
    It's unclear to me what purpose it serves to write $3^xtoinfty$. It makes the next expression, $frac{3^x}{4^x} to 0$, look wrong even though it actually is correct.
    $endgroup$
    – David K
    Jan 8 at 14:28
















  • 2




    $begingroup$
    $lim_{x to infty} left( 4^4+frac{3^x}{4^x} times 4^2 right)=lim_{x to infty}4^4 +lim_{x to infty}left(dfrac{3^x}{4^x}right)4^2$? While its true you might want to explain for rigor.
    $endgroup$
    – Yadati Kiran
    Dec 22 '18 at 4:26












  • $begingroup$
    It's unclear to me what purpose it serves to write $3^xtoinfty$. It makes the next expression, $frac{3^x}{4^x} to 0$, look wrong even though it actually is correct.
    $endgroup$
    – David K
    Jan 8 at 14:28










2




2




$begingroup$
$lim_{x to infty} left( 4^4+frac{3^x}{4^x} times 4^2 right)=lim_{x to infty}4^4 +lim_{x to infty}left(dfrac{3^x}{4^x}right)4^2$? While its true you might want to explain for rigor.
$endgroup$
– Yadati Kiran
Dec 22 '18 at 4:26






$begingroup$
$lim_{x to infty} left( 4^4+frac{3^x}{4^x} times 4^2 right)=lim_{x to infty}4^4 +lim_{x to infty}left(dfrac{3^x}{4^x}right)4^2$? While its true you might want to explain for rigor.
$endgroup$
– Yadati Kiran
Dec 22 '18 at 4:26














$begingroup$
It's unclear to me what purpose it serves to write $3^xtoinfty$. It makes the next expression, $frac{3^x}{4^x} to 0$, look wrong even though it actually is correct.
$endgroup$
– David K
Jan 8 at 14:28






$begingroup$
It's unclear to me what purpose it serves to write $3^xtoinfty$. It makes the next expression, $frac{3^x}{4^x} to 0$, look wrong even though it actually is correct.
$endgroup$
– David K
Jan 8 at 14:28












2 Answers
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$begingroup$

Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).



I will make a nitpick with one thing you said, though:




Since, as $x to infty, 3^x to infty, frac{3^x}{4^x} to 0$




Technically, since $3^x$ and $4^x$ both approach $infty$ as $x to infty$, then under your logic we obtain an indeterminate form:



$$frac{3^x}{4^x} to frac{infty}{infty}$$



It would be better to regroup $3^x/4^x$ as $(3/4)^x$. Then clearly, as $x to infty$, $(3/4)^x to 0$ because $3/4 < 1$. (If you're not convinced, notice if you take $x = 1, 2, 3, 4$ and so on that $(3/4)^x$ clearly is decreasing.)






share|cite|improve this answer









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    4












    $begingroup$

    Yup, the answer is correct.



    Remark:




    • While $3^x to infty$ is true, note that it is not used in the proof though.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      6












      $begingroup$

      Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).



      I will make a nitpick with one thing you said, though:




      Since, as $x to infty, 3^x to infty, frac{3^x}{4^x} to 0$




      Technically, since $3^x$ and $4^x$ both approach $infty$ as $x to infty$, then under your logic we obtain an indeterminate form:



      $$frac{3^x}{4^x} to frac{infty}{infty}$$



      It would be better to regroup $3^x/4^x$ as $(3/4)^x$. Then clearly, as $x to infty$, $(3/4)^x to 0$ because $3/4 < 1$. (If you're not convinced, notice if you take $x = 1, 2, 3, 4$ and so on that $(3/4)^x$ clearly is decreasing.)






      share|cite|improve this answer









      $endgroup$


















        6












        $begingroup$

        Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).



        I will make a nitpick with one thing you said, though:




        Since, as $x to infty, 3^x to infty, frac{3^x}{4^x} to 0$




        Technically, since $3^x$ and $4^x$ both approach $infty$ as $x to infty$, then under your logic we obtain an indeterminate form:



        $$frac{3^x}{4^x} to frac{infty}{infty}$$



        It would be better to regroup $3^x/4^x$ as $(3/4)^x$. Then clearly, as $x to infty$, $(3/4)^x to 0$ because $3/4 < 1$. (If you're not convinced, notice if you take $x = 1, 2, 3, 4$ and so on that $(3/4)^x$ clearly is decreasing.)






        share|cite|improve this answer









        $endgroup$
















          6












          6








          6





          $begingroup$

          Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).



          I will make a nitpick with one thing you said, though:




          Since, as $x to infty, 3^x to infty, frac{3^x}{4^x} to 0$




          Technically, since $3^x$ and $4^x$ both approach $infty$ as $x to infty$, then under your logic we obtain an indeterminate form:



          $$frac{3^x}{4^x} to frac{infty}{infty}$$



          It would be better to regroup $3^x/4^x$ as $(3/4)^x$. Then clearly, as $x to infty$, $(3/4)^x to 0$ because $3/4 < 1$. (If you're not convinced, notice if you take $x = 1, 2, 3, 4$ and so on that $(3/4)^x$ clearly is decreasing.)






          share|cite|improve this answer









          $endgroup$



          Yup! Your steps almost all follow logically and evaluate to the correct limit, so your solution is correct (almost).



          I will make a nitpick with one thing you said, though:




          Since, as $x to infty, 3^x to infty, frac{3^x}{4^x} to 0$




          Technically, since $3^x$ and $4^x$ both approach $infty$ as $x to infty$, then under your logic we obtain an indeterminate form:



          $$frac{3^x}{4^x} to frac{infty}{infty}$$



          It would be better to regroup $3^x/4^x$ as $(3/4)^x$. Then clearly, as $x to infty$, $(3/4)^x to 0$ because $3/4 < 1$. (If you're not convinced, notice if you take $x = 1, 2, 3, 4$ and so on that $(3/4)^x$ clearly is decreasing.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 4:29









          Eevee TrainerEevee Trainer

          6,0931936




          6,0931936























              4












              $begingroup$

              Yup, the answer is correct.



              Remark:




              • While $3^x to infty$ is true, note that it is not used in the proof though.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                Yup, the answer is correct.



                Remark:




                • While $3^x to infty$ is true, note that it is not used in the proof though.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Yup, the answer is correct.



                  Remark:




                  • While $3^x to infty$ is true, note that it is not used in the proof though.






                  share|cite|improve this answer









                  $endgroup$



                  Yup, the answer is correct.



                  Remark:




                  • While $3^x to infty$ is true, note that it is not used in the proof though.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 4:26









                  Siong Thye GohSiong Thye Goh

                  101k1466118




                  101k1466118






























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