- #1

FaroukYasser

- 62

- 3

## Homework Statement

Show that ##\lim _{ n\rightarrow \infty }{ \left( \frac { \sqrt { n+c } +d }{ \sqrt [ 3 ]{ { n }^{ 2 }+an+b } } \right) } =0,\quad n>-c ##

## Homework Equations

Sandwich theorem

## 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