Does $\lim_{x\to 0^+} \sqrt{x}$ exist or not? Does this right-handed limit exist or not? $$\lim_{x\to 0^+} \sqrt{x}$$ Schaum's Easy Outline of Calculus (Second Edition) says it does. And doesn't. An example in the book states: > The function $f(x)=\sqrt{x}$; then $f$ is defined only to the right of zero. Okay. So the right-handed limit _does_ exist. > Hence, $\lim_{x\to 0} \sqrt{x}=\lim_{x\to 0^+} \sqrt{x}=0$. Okay. I'm still with you. The limit is $0$. > Of course, $\lim_{x\to 0^+} \sqrt{x}$ does not exist,... Qué, Mr. Fawlty? > ... since $\sqrt{x}$ is not defined when $x<0$. Okay, so are they messing with me? Is my coffee too weak? Too strong? Is there some subtle truth about limits that escapes me? Or is that a typo? Did they mean in that last line to omit the '$+$' by the '$0$' and write "Of course, $\lim_{x\to 0} \sqrt{x}$ does not exist,..."? EDIT: I think a minus sign is intended instead of a plus sign in that last limit. ![Schaums limit](

Both $$ \lim_{x\to 0} \sqrt{x} \quad \text{and}\quad \lim_{x\to 0^+}\sqrt{x} $$ exist. In general for a function $f$ with domain $D(f)$, recall the definition of the $$ \lim_{x\to a} f(x) = L. $$ The definition says that this means that: For all $\epsilon >0$ there is a $\delta >0$ such that if $x\in D(f)$ and $0<\lvert x - a \rvert < \delta$ then $\lvert f(x) - L\rvert<\epsilon$. Often we don't write in the requirement that $x$ be in the domain of $f$, but this is a requirement.

Likewise the right hand limit exists.

See this Wikipedia article for more on this: <