A long limit

Let $f(y)$ be that seventh root. We have that $f(y) - 2y = \frac{f(y)^7-128y^7}{f(y)^6 + (2y)f(y)^5 + \cdots + (2y)^6}.$ The top of this fraction is a polynomial of degree $6$ , equal to $1488 y^6 + c_5 y^5 + \cdots + c_0$ . Divide the top and bottom by $y^6$ to get $\frac{1488 + c_5/y + \cdots + c_0/y^6}{(f(y)/y)^6 + 2(f(y)/y)^5 + \cdots + 2^6}.$

Since $\lim\limits_{y\to\infty} \frac{f(y)}{y} = 2$ , the limit we want is $\frac{1488}{2^6 + 2^6 + \cdots 2^6} = \frac{1488}{448} = \frac{93}{28}$ . So the answer is $\fbox{121}$ .

The slope of the asymptote of the radical

$\sqrt [ 7 ]{ (8y+3)(y+1)(y+2)(y+3)(2y+3)(2y+5)(4y+5) }$

is $2y$ , because the highest order term in the radical (when expanded) is $128{ y }^{ 7 }$ . Hence, this is cancelled by the subtraction of $2y$ from the radical, leaving a limit value which is the same as the $y$ intercept of the asymptote. Let $a$ be that $y$ intercept. Then the coefficient of the $2nd$ term of both the expanded polynomial in the radical and ${ (2y+a) }^{ 7 }$ has to be the same. Thus $a$ has to be

$a=\dfrac { 128(\dfrac { 3 }{ 8 } +\dfrac { 1 }{ 1 } +\dfrac { 2 }{ 1 } +\dfrac { 3 }{ 1 } +\dfrac { 5 }{ 2 } +\dfrac { 5 }{ 4 } ) }{ 128(7)(\dfrac { 1 }{ 2 } ) } =\dfrac { 93 }{ 28 } =\dfrac { A }{ B }$

A weird solution ----

We need to find

$L=\displaystyle\lim_{y \rightarrow \infty} \sqrt[7]{(8y+3)(y+1)(y+2)(y+3)(2y+3)(2y+5)(4y+5)}-2y$ .

We can rewrite it as

$L=\displaystyle\lim_{y \rightarrow \infty} 2 \times \sqrt[7]{\left(y+\frac{3}{8}\right)(y+1)(y+2)(y+3)\left(y+\frac{3}{2}\right)\left(y+\frac{5}{2}\right)\left(y+\frac{5}{4}\right)}-2y$

Now, consider the terms $\left(y+\frac{3}{8}\right), (y+1), (y+2), (y+3), \left(y+\frac{3}{2}\right), \left(y+\frac{5}{2}\right)$ and $\left(y+\frac{5}{4}\right)$ .

As $y \rightarrow \infty$ , all these terms become equal.

We can check this by dividing each of the above terms by every other term and then applying $\displaystyle\lim_{y \rightarrow \infty}$ to it.

Hence, we can apply the fact that for equal terms, $A.M=G.M$ .

Doing so,

$\displaystyle\lim_{y \rightarrow \infty} \sqrt[7]{\left(y+\frac{3}{8}\right)(y+1)(y+2)(y+3)\left(y+\frac{3}{2}\right)\left(y+\frac{5}{2}\right)\left(y+\frac{5}{4}\right)}=\displaystyle\lim_{y \rightarrow \infty} \dfrac{\left(y+\frac{3}{8}\right)+(y+1)+(y+2)+(y+3)+\left(y+\frac{3}{2}\right)+\left(y+\frac{5}{2}\right)+\left(y+\frac{5}{4}\right)}{7}$ .

Hence,

$L= \displaystyle\lim_{y \rightarrow \infty} 2\left(\dfrac{93}{56}+y\right)-2y$

$\boxed{L=\dfrac{93}{28}}$

$\therefore \boxed{A+B=121}$

Take out factor of $2y$ and use Binomial Expansion on each of the sevenths roots.

Solving the problem by using CAUCHY's INEQUALITY:

$L\quad =\quad \lim _{ y\rightarrow \infty }{ [\sqrt [ 7 ]{ (8y+3)(y+1)(y+2)(y+3)(2y+3)(2y+5)(4y+5) } -2y] } \quad =\quad \\ 2\lim _{ y\rightarrow \infty }{ [\sqrt [ 7 ]{ (y+3/8)(y+1)(y+2)(y+3)(y+3/2)(y+5/2)(y+5/4) } -y } ]$

by Cauchy's inequality:

$\sqrt [ 7 ]{ (y+3/8)(y+1)(y+2)(y+3)(y+3/2)(y+5/2)(y+5/4) } \le \\ \le \quad \frac { (y+3/8)+(y+1)+(y+2)+(y+3)+(y+3/2)+(y+5/2)+(y+5/4) }{ 7 } =y+\frac { 93 }{ 56 }$

where equality holds for $y+3/8\approx y+1\approx y+2\approx y+3\approx y+3/2\approx y+5/2\approx y+5/4$ it works if $y\rightarrow \infty$

$\Longrightarrow \lim _{ y\rightarrow \infty }{ [\sqrt [ 7 ]{ (y+3/8)(y+1)(y+2)(y+3)(y+3/2)(y+5/2)(y+5/4) } \quad =\quad } y+\frac { 93 }{ 56 }$

$\Longrightarrow \quad L\quad =\quad 2(y+\frac { 93 }{ 56 } -y)=\frac { 93 }{ 28 } =\frac { A }{ B } \quad \quad \quad \quad \Longrightarrow \quad Answer\quad is\quad A+B=121$