Limit $lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$











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I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.



$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$










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  • Are we allowed to use L'Hospital Rule?
    – Sujit Bhattacharyya
    Dec 1 at 12:48










  • I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
    – Michael
    Dec 1 at 12:50






  • 1




    Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
    – Michael
    Dec 1 at 13:02















up vote
0
down vote

favorite












I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.



$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$










share|cite|improve this question









New contributor




Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Are we allowed to use L'Hospital Rule?
    – Sujit Bhattacharyya
    Dec 1 at 12:48










  • I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
    – Michael
    Dec 1 at 12:50






  • 1




    Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
    – Michael
    Dec 1 at 13:02













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.



$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$










share|cite|improve this question









New contributor




Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I really don't know how to evaluate the limit of the following function. That's the only exercise with logarithms I had so I don`t really have enough experience to know how to evaluate the limit.



$$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)$$







limits






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edited Dec 1 at 13:00









Arjang

5,55162363




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asked Dec 1 at 12:45









Michael

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23




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Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • Are we allowed to use L'Hospital Rule?
    – Sujit Bhattacharyya
    Dec 1 at 12:48










  • I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
    – Michael
    Dec 1 at 12:50






  • 1




    Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
    – Michael
    Dec 1 at 13:02


















  • Are we allowed to use L'Hospital Rule?
    – Sujit Bhattacharyya
    Dec 1 at 12:48










  • I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
    – Michael
    Dec 1 at 12:50






  • 1




    Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
    – Michael
    Dec 1 at 13:02
















Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48




Are we allowed to use L'Hospital Rule?
– Sujit Bhattacharyya
Dec 1 at 12:48












I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50




I've been thinking about L'Hospital Rule but this exercise is from the lesson number 4 when L'Hospital Rule is being done during lesson number 5. I'm pretty sure there is another way.
– Michael
Dec 1 at 12:50




1




1




Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02




Thanks for all the answers. I really missed the part that ln(1+x)∼x. Now everything will be much easier.
– Michael
Dec 1 at 13:02










7 Answers
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accepted










Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
$$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
therefore
$$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$






share|cite|improve this answer





















  • It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
    – gimusi
    Dec 1 at 13:47












  • Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
    – gimusi
    Dec 1 at 13:49




















up vote
1
down vote













Since $$log(1+t) sim t$$ as $t to 0$, we have that



$$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$






share|cite|improve this answer




























    up vote
    1
    down vote













    HINT



    $$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$






    share|cite|improve this answer




























      up vote
      1
      down vote













      As an alternative way, you can also note that



      $$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$



      $$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$



      and



      $$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$






      share|cite|improve this answer




























        up vote
        1
        down vote













        First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.



        Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
        $$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$






        share|cite|improve this answer



















        • 2




          It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
          – Botond
          Dec 1 at 13:09










        • Thanks for pointing that. I edited.
          – Sujit Bhattacharyya
          Dec 1 at 13:18


















        up vote
        1
        down vote













        $lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$



        $log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$



        $log left(dfrac{e^7}{e^2}right)= 5.$



        Used :



        $lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.






        share|cite|improve this answer






























          up vote
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          down vote













          $displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.



          $$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$



          Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.



          Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!






          share|cite|improve this answer





















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            7 Answers
            7






            active

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            7 Answers
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            up vote
            1
            down vote



            accepted










            Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
            $$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
            therefore
            $$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$






            share|cite|improve this answer





















            • It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
              – gimusi
              Dec 1 at 13:47












            • Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
              – gimusi
              Dec 1 at 13:49

















            up vote
            1
            down vote



            accepted










            Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
            $$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
            therefore
            $$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$






            share|cite|improve this answer





















            • It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
              – gimusi
              Dec 1 at 13:47












            • Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
              – gimusi
              Dec 1 at 13:49















            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
            $$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
            therefore
            $$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$






            share|cite|improve this answer












            Use equivalents: we know that, near $0$, $ln(1+u)sim u$, so
            $$lnbiggl(frac{x+7}{x+2}biggr)=lnbiggl(1+frac{5}{x+2}biggr)sim_{xto +infty}frac{5}{x+2},$$
            therefore
            $$xlnbiggl(frac{x+7}{x+2}biggr)sim_{xto +infty}frac{5x}{x+2}to 5.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 at 12:55









            Bernard

            116k637108




            116k637108












            • It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
              – gimusi
              Dec 1 at 13:47












            • Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
              – gimusi
              Dec 1 at 13:49




















            • It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
              – gimusi
              Dec 1 at 13:47












            • Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
              – gimusi
              Dec 1 at 13:49


















            It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
            – gimusi
            Dec 1 at 13:47






            It should always recall in my opinion that equivalents need to be used with great attention since in some cases their use by not expert could lead to big mistakes. It would be better use the form by little-o or big-O notation in general. In that case it is of course fine (+1).
            – gimusi
            Dec 1 at 13:47














            Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
            – gimusi
            Dec 1 at 13:49






            Simple example since $cos xsim 1$ we could erroneously conclude that $(1-cos x)/x^2sim 0$ which is wrong of course.
            – gimusi
            Dec 1 at 13:49












            up vote
            1
            down vote













            Since $$log(1+t) sim t$$ as $t to 0$, we have that



            $$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$






            share|cite|improve this answer

























              up vote
              1
              down vote













              Since $$log(1+t) sim t$$ as $t to 0$, we have that



              $$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Since $$log(1+t) sim t$$ as $t to 0$, we have that



                $$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$






                share|cite|improve this answer












                Since $$log(1+t) sim t$$ as $t to 0$, we have that



                $$lim_{x to infty} x log left(1 + frac 5{x+2}right) = lim_{x to infty} xcdot frac 5{x+2} = 5$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 at 12:54









                Ant

                17.3k22873




                17.3k22873






















                    up vote
                    1
                    down vote













                    HINT



                    $$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$






                    share|cite|improve this answer

























                      up vote
                      1
                      down vote













                      HINT



                      $$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$






                      share|cite|improve this answer























                        up vote
                        1
                        down vote










                        up vote
                        1
                        down vote









                        HINT



                        $$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$






                        share|cite|improve this answer












                        HINT



                        $$xlogleft(frac{x+7}{x+2}right)=log left(1 + frac 5{x+2}right)^x$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Dec 1 at 12:56









                        gimusi

                        89.7k74495




                        89.7k74495






















                            up vote
                            1
                            down vote













                            As an alternative way, you can also note that



                            $$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$



                            $$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$



                            and



                            $$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$






                            share|cite|improve this answer

























                              up vote
                              1
                              down vote













                              As an alternative way, you can also note that



                              $$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$



                              $$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$



                              and



                              $$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$






                              share|cite|improve this answer























                                up vote
                                1
                                down vote










                                up vote
                                1
                                down vote









                                As an alternative way, you can also note that



                                $$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$



                                $$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$



                                and



                                $$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$






                                share|cite|improve this answer












                                As an alternative way, you can also note that



                                $$xlnbigg(frac{x+7}{x+2}bigg) = lnbigg(frac{x+7}{x+2}bigg)^x = lnbigg(1+frac{5}{x+2}bigg)^x$$



                                $$= lnBiggl[bigg(1+frac{5}{x+2}bigg)^{x+2}Biggl]^{frac{x}{x+2}}$$



                                and



                                $$lim_{nto infty} bigg(1+frac{x}{n}bigg)^n = e^x$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Dec 1 at 13:09









                                KM101

                                3,273417




                                3,273417






















                                    up vote
                                    1
                                    down vote













                                    First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.



                                    Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
                                    $$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$






                                    share|cite|improve this answer



















                                    • 2




                                      It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
                                      – Botond
                                      Dec 1 at 13:09










                                    • Thanks for pointing that. I edited.
                                      – Sujit Bhattacharyya
                                      Dec 1 at 13:18















                                    up vote
                                    1
                                    down vote













                                    First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.



                                    Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
                                    $$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$






                                    share|cite|improve this answer



















                                    • 2




                                      It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
                                      – Botond
                                      Dec 1 at 13:09










                                    • Thanks for pointing that. I edited.
                                      – Sujit Bhattacharyya
                                      Dec 1 at 13:18













                                    up vote
                                    1
                                    down vote










                                    up vote
                                    1
                                    down vote









                                    First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.



                                    Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
                                    $$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$






                                    share|cite|improve this answer














                                    First note that, $$lim_{xto 0}frac{log(1+mx)}{x}=m$$ for any real $m$.



                                    Putting, $x=1/y$ we get, $xtoinftyimplies yto 0$ and the limit becomes,
                                    $$lim_{xtoinfty}xlogleft(frac{x+7}{x+2}right)=lim_{yto0}frac{1}{y}logleft(frac{1+7y}{1+2y}right)=lim_{yto0}frac{1}{y}{log(1+7y)-log(1+2y)}=7-2=5$$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Dec 1 at 13:18

























                                    answered Dec 1 at 12:56









                                    Sujit Bhattacharyya

                                    940316




                                    940316








                                    • 2




                                      It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
                                      – Botond
                                      Dec 1 at 13:09










                                    • Thanks for pointing that. I edited.
                                      – Sujit Bhattacharyya
                                      Dec 1 at 13:18














                                    • 2




                                      It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
                                      – Botond
                                      Dec 1 at 13:09










                                    • Thanks for pointing that. I edited.
                                      – Sujit Bhattacharyya
                                      Dec 1 at 13:18








                                    2




                                    2




                                    It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
                                    – Botond
                                    Dec 1 at 13:09




                                    It's trivially true of $m=0$ as well, because of $frac{0}{x} to 0$.
                                    – Botond
                                    Dec 1 at 13:09












                                    Thanks for pointing that. I edited.
                                    – Sujit Bhattacharyya
                                    Dec 1 at 13:18




                                    Thanks for pointing that. I edited.
                                    – Sujit Bhattacharyya
                                    Dec 1 at 13:18










                                    up vote
                                    1
                                    down vote













                                    $lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$



                                    $log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$



                                    $log left(dfrac{e^7}{e^2}right)= 5.$



                                    Used :



                                    $lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.






                                    share|cite|improve this answer



























                                      up vote
                                      1
                                      down vote













                                      $lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$



                                      $log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$



                                      $log left(dfrac{e^7}{e^2}right)= 5.$



                                      Used :



                                      $lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.






                                      share|cite|improve this answer

























                                        up vote
                                        1
                                        down vote










                                        up vote
                                        1
                                        down vote









                                        $lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$



                                        $log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$



                                        $log left(dfrac{e^7}{e^2}right)= 5.$



                                        Used :



                                        $lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.






                                        share|cite|improve this answer














                                        $lim_{x rightarrow infty} log(dfrac{(1+7/x)^x}{(1+2/x)^x})=$



                                        $log dfrac{lim_{x rightarrow infty}(1+7/x)^x}{lim_{x rightarrow infty}(1+2/x)^x}=$



                                        $log left(dfrac{e^7}{e^2}right)= 5.$



                                        Used :



                                        $lim_{x rightarrow infty} (1+a/x)^x=e^a$, $a$ real.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Dec 1 at 13:45

























                                        answered Dec 1 at 13:39









                                        Peter Szilas

                                        10.4k2720




                                        10.4k2720






















                                            up vote
                                            0
                                            down vote













                                            $displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.



                                            $$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$



                                            Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.



                                            Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!






                                            share|cite|improve this answer

























                                              up vote
                                              0
                                              down vote













                                              $displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.



                                              $$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$



                                              Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.



                                              Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!






                                              share|cite|improve this answer























                                                up vote
                                                0
                                                down vote










                                                up vote
                                                0
                                                down vote









                                                $displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.



                                                $$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$



                                                Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.



                                                Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!






                                                share|cite|improve this answer












                                                $displaystyle lim_{x to infty} x logbiggr(frac{x+7}{x+2}biggl)$ is of the indeterminate form $0 cdot infty$. We can rewrite the expression to be of the form $dfrac{infty}{frac{1}{0}}$ to be able to apply L'Hopital's Rule.



                                                $$x log biggr(frac{x+7}{x+2}biggl)= dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} implies displaystyle lim_{x to infty} dfrac{log (frac{x+7}{x+2})}{frac{1}{x}} = displaystyle lim_{x to infty} dfrac{frac{x+2}{x+7}}{frac{-1}{x^2}}cdot frac{d}{dx}biggr(frac{x+7}{x+2}biggl)$$



                                                Thus the limit translates to the new equivalent simplified expression i.e. $displaystyle lim_{x to infty} dfrac{5x^2}{(x+7)(x+2)}$.



                                                Therefore, $displaystyle lim_{x to infty} x log biggr(frac{x+7}{x+2}biggl)=5$. Cheers!







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                                                Paras Khosla

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