How can I determine limit: $\lim\limits_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$? How can I determine limit:
$$\lim_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$$
Thanks
 A: If you are unfamiliar with L'Hopital's rule, this is what you can do.
$$ \lim_{x \rightarrow 4} \frac{ x - 4 } { x - \sqrt{x} - 2} = \lim_{x \rightarrow 4} \frac{ (\sqrt{x} - 2) ( \sqrt{x} + 2) } { (\sqrt{x} -2) ( \sqrt{x} +1) }=  \lim_{x \rightarrow 4} \frac{ ( \sqrt{x} + 2) } { ( \sqrt{x} +1) } = \frac{ 4}{3}.$$
A: Note that the denominator tends to $4-\sqrt{4-2} = 4-\sqrt2$ as $x \to 4$ and the numerator tends to $0$ as $x \to 4$.
EDIT
The answer to the new question is as follows.
We have
$$\dfrac{x-4}{x-\sqrt{x}-2} = \dfrac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)} = \dfrac{2+\sqrt{x}}{1+\sqrt{x}} \,\,\,\,\, \forall x \neq 4$$
Now take the limit as $x \to 4$.
A: Write $t=\sqrt x$, now we can write the limit as:
$$\lim_{t\to 2}\frac{t^2-4}{t^2-t-2}=\lim_{t\to2}\frac{(t-2)(t+2)}{(t-2)(t+1)}=\lim_{t\to 2}\frac{t+2}{t+1}=\frac{2+2}{2+1}=\frac43$$
A: Let $h=x-4$ then
$$\frac{x-4}{x-\sqrt[]{x}-2}=\frac{h}{h-\sqrt[]{h+4}+2}=\frac{h}{h-2\sqrt[]{1+h/4}+2}\sim_0\frac{h}{h-2(1+h/8)+2}=\frac{4}{3}$$
A: The post has been edited, now the solution is easy if you apply L'Hopital rule:
$$\lim_{x \to 4} \frac{x - 4}{x - \sqrt{x} - 2} = \lim_{x \to 4}\frac{1}{1 - \frac{1}{2\sqrt{x}}} = \frac{1}{1 - \frac{1}{2\sqrt{4}}} = \frac{4}{3}$$
