Question

X -> ∞ * (sqrt(x - a) - sqrt(bx))​

Answer 1

Simplify the limand in the following way.

$\displaystyle \lim_{x\to\infty} \left(\sqrt{x-a} - \sqrt{bx}\right) = \lim_{x\to\infty} \dfrac{\left(\sqrt{x-a}\right)^2 - \left(\sqrt{bx}\right)^2}{\sqrt{x-a} + \sqrt{bx}} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \lim_{x\to\infty} \frac{(x-a) - bx}{\sqrt x \left(\sqrt{1-\frac ax} + \sqrt b\right)} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \lim_{x\to\infty} \frac{(1-b)\sqrt x - \frac a{\sqrt x}}{\sqrt{1 - \frac ax} + \sqrt b}$

Now,

$\displaystyle \lim_{x\to\infty} \frac a{\sqrt x} = 0$

$\displaystyle \lim_{x\to\infty} \sqrt{1-\frac ax} = \sqrt{1-\lim_{x\to\infty}\frac ax}} = \sqrt1 = 1$

$\implies \displaystyle \lim_{x\to\infty} \left(\sqrt{x-a} - \sqrt{bx}\right) = \frac{1-b}{\sqrt b} \lim_{x\to\infty} \sqrt x$

and therefore

$\displaystyle \lim_{x\to\infty} \left(\sqrt{x-a} - \sqrt{bx}\right) = \begin{cases} 0 & \text{if } b = 1 \\ -\infty & \text{if } b > 1\end{cases}$

and does not exist otherwise.