# Is there an easier way to find this limit rigorously?

1. Sep 12, 2015

### FaroukYasser

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.

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

2. Sep 12, 2015

### Staff: Mentor

Divide both numerator and denominator by the same right power of n, and you can take easy limits everywhere.

3. Sep 12, 2015

### FaroukYasser

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?

4. Sep 12, 2015

### 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.

5. Sep 12, 2015

### Ray Vickson

There is an easier way: write the numerator as
$$\text{numerator} = \sqrt{n+c} + d = \sqrt{n} \left( \left(1 + \frac{c}{n} \right)^{1/2} + \frac{d}{\sqrt{n}} \right)$$
and the denominator as
$$\text{denominator} = \sqrt[3]{n^2 + an + b} = n^{2/3} \left( 1 + \frac{a}{n} + \frac{b}{n^2} \right)^{1/3}$$
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.

6. Sep 12, 2015

### FaroukYasser

Thanks a lot!