Solving a limit of radicals without l'Hopital

Question

I'm struggling with the following limit. I have tried to manipulate it in a number of ways, each resulting in a dead-end or circling back to the original form. Any hints on how to solve this limit would be appreciated.

$$\lim _{x\rightarrow 2}\dfrac {\sqrt{\dfrac {2}{x}}-1}{2-x}$$

Answer 1

Hint:

$$ \dfrac {\sqrt{\dfrac {2}{x}}-1}{2-x}=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}(\sqrt{2}-\sqrt{x})(\sqrt{2}+\sqrt{x})} $$

Answer 2

"Solving" is the wrong word. "Evaluating" is the right one. One solves an equation for a variable; one solves a problem, and you could speak of solving the problem of evaluating the limit. One does not "solve" an expression; one evaluates it or simplifies it or does other things with it.

Rationalizing the numerator works in this case: \begin{align} & \frac{\left( \sqrt{\frac 2 x } -1 \right)\left( \sqrt{\frac 2 x} + 1\, \right)}{(2-x)\left( \sqrt{\frac 2 x} + 1\, \right)} = \frac{\frac 2 x -1 }{(2-x)\left( \sqrt{\frac 2 x} + 1\, \right)} \\[12pt] = {} & \frac{2-x}{x(2-x)\left( \sqrt{\frac 2 x} + 1\, \right)} = \frac 1 {x \left( \sqrt{\frac 2 x} + 1\, \right)} \end{align}

It is easy to find the limit of that last exprssion as $x\to 2$.

Answer 3

Hint:

Let $$ \sqrt{\dfrac2x}-1=h$$

$\implies x=\dfrac2{(1+h)^2}$ and $h\to0$