Can the difference of 2 undefined limits be defined? Is this limit defined or undefined?
$$\lim\limits_{x \to 0+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)$$
When I apply the rule of difference of limits, it's undefined. But, when I manipulate it, it gives me zero. And the graph of the function indicates it's defined on the right side.
By multiplying by $\frac{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$:
$$\lim\limits_{x \to 0+} \frac{\left( \sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}} \, \right) \left(\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}} \, \right)}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
$$=\lim\limits_{x \to 0+} \frac{\frac{1}{x}+2-\frac{1}{x}}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
$$=\lim\limits_{x \to 0+} \frac{2}{\sqrt{\frac{1}{x}+2}+\sqrt{\frac{1}{x}}}$$
Then, we multiply by $\frac{\sqrt{x}}{\sqrt{x}}$:
$$=\lim\limits_{x \to 0} \frac{2\sqrt{x}}{\sqrt{1+2x}+1}$$
And, we substitute:
$$=\frac{2\sqrt{0}}{\sqrt{1+2\times0}+1} = 0$$
So, is this limit defined or not? and what's my error, if any?

You write $x\to 0+$ but it's often written as $x\to 0^+$. You asked if your limit is defined or undefined. I'll answer that with another two ways.
$$\lim_{x\to 0^+} \left(\sqrt{\frac{1}{x}+2}-\sqrt{\frac{1}{x}}\right)=$$
$$=\lim_{x\to 0^+}\left(\frac{\sqrt{1+2x}-1}{x}\cdot \sqrt{x}\right)=$$
Use the definition of a derivative and a limit multiplication rule/law.
$$=(\sqrt{1+2x})\bigg|_{x=0^+}\cdot 0=0,$$
because the derivative is a real number.
Another way is using Newton's generalized binomial theorem, also see Binomial series, which converges when $|x|<1$ but it can diverge when $|x|\ge 1$, see the Binomial series link for more information. $$\lim_{x\to 0^+}\frac{\sqrt{1+2x}-1}{x}=$$
$$=\lim_{x\to 0^+}\frac{(1+2x)^{\frac{1}{2}}-1}{x}=$$
$$=\lim_{x\to 0^+}\frac{1+\frac{1}{2}2x+o(x)-1}{x}=$$
$$=\lim_{x\to 0^+}\left(1+\frac{o(x)}{x}\right)=1+0=1$$
