Let f:[1,3]R be a function satisfying x[x] f(x)√6-x for all x2, and f(2)=1, where [x] is t…

Question

Let $f:[1,3]	o\mathbb{R}$ be a function satisfying $\dfrac{x}{[x]}\le f(x)\le\sqrt{6-x}$ for all $x
e2$, and $f(2)=1$, where $[x]$ is the greatest integer $\le x$.

Statement 1: $\displaystyle\lim_{x	o2^-}f(x)$ exists.

Statement 2: $f$ is continuous at $x=2$.

Accepted Answer

For $x	o2^-$, $[x]=1$, so $\dfrac x{[x]}	o2$ and $\sqrt{6-x}	o2$; by the Sandwich theorem, $\displaystyle\lim_{x	o2^-}f(x)=2$ --- Statement 1 true.

For $x	o2^+$, $[x]=2$, so $\dfrac x{[x]}	o1$ and $\sqrt{6-x}	o2$; the Sandwich theorem does not pin down a unique value, so $\displaystyle\lim_{x	o2^+}f(x)$ need not exist/match. Since $f(2)=1$ but LHL $=2$, $f$ is not continuous at $x=2$ --- Statement 2 false.

Solutions · 1