If $~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ $, find $~ a+b=~?$ If $$~\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32~ ,$$ find $~ a+b=~?$
I get $a-16b = 24$,
But how to get $a+b$?
a,b is positive integer
 A: $$
\sqrt{64x^2 + ax + 7} =\sqrt{(8x)^2 + ax + 7} =\sqrt{(8x)^2 + 2\times(?)\times(8x)+ 7} \\
$$note that $$2\times(?)\times(8x)=ax \to ?=\frac{a}{16}$$so
$$\sqrt{(8x)^2 + 2(\frac{a}{16})(8x)+ 7}=\sqrt{(8x+\frac{a}{16})^2 -(\frac{a}{16})^2+ 7}\\x\to \infty \sqrt{(8x+\frac{a}{16})^2 -(\frac{a}{16})^2+ 7}\sim |8x+\frac{a}{16}| $$ 
so 
$$lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac{3}{2}\\
lim_{x\to\infty}|8x+\frac{a}{16}| - 8x + b = \frac{3}{2}$$ now can you take over? 
a,b are a positive integer so 
$$\frac{a}{16}+b=\frac{3}{2}\\b=0\to \frac{a}{16}+0=\frac{3}{2} \to a=24 \to a+b=24\\b=1 \to \frac{a}{16}+1=\frac{3}{2} \to a=8 \to a+b=1+8=9\\b=2,3,... impossible $$
A: $\lim_{x\to\infty}\sqrt{64x^2 + ax + 7} - 8x + b = \frac32$
Define $k(x) \equiv 
\sqrt{64x^2 + ax + 7} - 8x$ 
$k(x)= (\sqrt{64x^2 + ax + 7} - 8x) \frac{\sqrt{64x^2 + ax + 7} + 8x}{\sqrt{64x^2 + ax + 7} + 8x}$
$$=\frac{64x^2 + ax + 7 - 64x^2}{\sqrt{64x^2 + ax + 7} + 8x}$$
$$=\frac{a + \frac7x}{\sqrt{64 + \frac ax + \frac 7 {x^2}} + 8}$$
So $\lim_{x\rightarrow\infty}k(x) = \frac a {16}$
And by assumption, $\lim_{x\rightarrow \infty} k(x) + b = \frac 3 2$, so 
$$\frac a {16} + b = \frac 3 2$$
This is a necessary and sufficient condition for the conditions stated, meaning there are many possible solutions. For example, setting $a = 0, b = \frac 3 2$ works and then $a + b = \frac 3 2$.
