Is there an easier way to find this limit rigorously?

[FaroukYasser]

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

 

[mfb]

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

 

[FaroukYasser]

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

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 want to make sure the steps are moving logically. And dividing the numerator and denominator by n^(2/3) would do the trick right?

 

[mfb]

And dividing the numerator and denominator by n^(2/3) would do the trick right?

Yes.

is my method ok or does it have any flaw in the logic?

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.

 

[Ray Vickson]

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

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.

 

[FaroukYasser]

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.

Thanks a lot!