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?
Best Answer
Remember that the rule that you referred to, "the rule of difference of limits", is not just the equation $$ \lim_{x\to a}(f(x)-g(x))=\lim_{x\to a}f(x)-\lim_{x\to a}g(x) $$ but rather the statement that, if both of the limits on the right side of this equation are real numbers, then the limit on the left side (is also a real number and) is given by this equation. So this rule does not apply to the limit in your question.
More generally, when learning rules (or theorems or principles or whatever they may be called), don't just learn formulas, but pay attention also to the words around them. The words are not just decoration but are essential for the correctness of the rule.