Question

Let $f\left(x\right)=(x+1)2^{-(\frac{1}{\left[x\right]}+\frac{1}{x})}$ and $f\left(0\right)=0$

• A. $f$ is continuous at x = 0
• B. $\underset{x\to 0^{+}}{lim}f\left(x\right)$ exists
• C. $\underset{x\to 0^{+}}{lim}f\left(x\right)$ does not exist
• D. $\underset{x\to 0}{\overset{+}{lim}}f\left(x\right)\ne \underset{x\to 0^{-}}{lim}f\left(x\right)$

Answer 1

The correct option is C $\underset{x\to 0^{+}}{lim}f\left(x\right)$ does not exist

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(x+1)2^{-(\frac{1}{\left[x\right]}+\frac{1}{x})}=(0+1)2^{-(\frac{1}{-1}+\frac{1}{0})}=2^{-\infty }\to 0$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}(x+1)2^{-(\frac{1}{\left[x\right]}+\frac{1}{x})}=\underset{x\to 0^{+}}{lim}(x+1)2^{-(\frac{1}{0}+\frac{1}{x})}$

The left-hand limit exists, and the value of the function tends to zero as the value of $x$ tends to zero.
The right-hand limit does not exist, because the function is not defined for $xϵ(0,1)$, where $\left[x\right]=0$.
So the correct answer is option C