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.
Best Answer
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: https://en.wikipedia.org/wiki/%28%CE%B5,_%CE%B4%29-definition_of_limit#Precise_statement