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Is there an easier way to find this limit rigorously?

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that ##\lim _{ n\rightarrow \infty }{ \left( \frac { \sqrt { n+c } +d }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \right) } =0,\quad n>-c ##

    2. Relevant equations
    Sandwich theorem

    3. The attempt at a solution

    Ok, So I know my method is extremely long, I'm just wandering if 1) It is correct and 2)If there is any better way than this.

    ##If\quad n>\frac { -b }{ a } And\quad n>{ d }^{ 2 }-c\\ \Longrightarrow \quad 0\quad <\quad \frac { \sqrt { n+c } +d }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \le \frac { \sqrt { n+c } +\left| d \right| }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \quad <\quad \frac { \sqrt { n+c } +\left| d \right| }{ \sqrt { { n }^{ 2 }+an+b } } <\frac { \sqrt { n+c } +\left| d \right| }{ \sqrt { { n }^{ 2 } } } =\frac { \sqrt { n+c } +\left| d \right| }{ n } \\ \\ <\quad \frac { \sqrt { n+c } +\left| d \right| \sqrt { n+c } }{ n } =\frac { (1+\left| d \right| )(\sqrt { n+c } ) }{ n } <\frac { (1+\left| d \right| )(\sqrt { n+n } ) }{ n } ,\quad for\quad n>\left| c \right| =\frac { \sqrt { 2 } (1+\left| d \right| ) }{ \sqrt { n } } \\ \therefore \quad for\quad n>max\left\{ \frac { -b }{ a } ,\left| c \right| ,{ d }^{ 2 }-c \right\} \quad \Longrightarrow \quad 0\quad <\quad \frac { \sqrt { n+c } +d }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \quad <\quad \frac { \sqrt { 2 } (1+\left| d \right| ) }{ \sqrt { n } } \\ \because \quad \lim _{ n\rightarrow \infty }{ \left( \frac { \sqrt { 2 } (1+\left| d \right| ) }{ \sqrt { n } } \right) } =0\quad Then\quad by\quad the\quad sandwitch\quad theorem\quad \lim _{ n\rightarrow \infty }{ \left( \frac { \sqrt { n+c } +d }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \right) } =0##

    Thanks in advance
     
  2. jcsd
  3. Sep 12, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Divide both numerator and denominator by the same right power of n, and you can take easy limits everywhere.
     
  4. Sep 12, 2015 #3
    Thanks. I was wandering though, is my method ok or does it have any flaw in the logic? I am just trying to exercise with the sandwich theorem so I just wanna make sure the steps are moving logically. And dividing the numerator and denominator by n^(2/3) would do the trick right?
     
  5. Sep 12, 2015 #4

    mfb

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    2016 Award

    Staff: Mentor

    Yes.
    That step does not work, you increase the denominator (in general), so you reduce the fraction when going from the left to the right:
    $$\frac { \sqrt { n+c } +\left| d \right| }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \quad <\quad \frac { \sqrt { n+c } +\left| d \right| }{ \sqrt { { n }^{ 2 }+an+b } }$$
    There is a long way, but it is complicated.
     
  6. Sep 12, 2015 #5

    Ray Vickson

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    Science Advisor
    Homework Helper

    There is an easier way: write the numerator as
    [tex] \text{numerator} = \sqrt{n+c} + d = \sqrt{n} \left( \left(1 + \frac{c}{n} \right)^{1/2} + \frac{d}{\sqrt{n}} \right) [/tex]
    and the denominator as
    [tex] \text{denominator} = \sqrt[3]{n^2 + an + b} = n^{2/3} \left( 1 + \frac{a}{n} + \frac{b}{n^2} \right)^{1/3} [/tex]
    If you really insist on using the sandwich theorem you could start by finding simple upper and lower bounds on ##(1 + x)^{1/2}## and ##(1+x)^{1/3}## for small ##|x|##. However, avoiding the sandwich theorem altogether seems much simpler.
     
  7. Sep 12, 2015 #6
    Thanks a lot!
     
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