On continuity of floor, modulus and fractional part function.

by mnulb   Last Updated May 16, 2018 14:20 PM

If $$f(x)=\begin{cases}\dfrac{e^{\lfloor x\rfloor}+|x|-1}{\lfloor x\rfloor+\{2x\}}&,\ x\neq0 \\\ \frac{1}{2}&,\ x=0\end{cases}$$comment on continuity of $f(x)$ at $x=0$. Where $\lfloor .\rfloor$ denotes floor function, $|.|$ denotes absolute value function and $\{.\}$ denotes fraction part.

As $x\rightarrow0$ we should calculate two-sided limits separately.

$\lim_{x\rightarrow 0^{-}}\dfrac{e^{\lfloor x\rfloor}+|x|-1}{\lfloor x\rfloor+\{2x\}}=\lim_{x\rightarrow 0^{-}}\dfrac{e^{-1}+0-1}{-1-2x}=\lim_{x\rightarrow 0^{-}}\dfrac{e^{-1}-1}{-1}=\dfrac{e-1}{e}\rightarrow{(*)}$

Now my doubt arises for $(*)$:

As $x\rightarrow{0^{-}}$, mean to say $x$ is roaming around really tiny negative quantities hence I wrote $\{2x\}$ as $-2x$ then again neglected $-2x$ in $-1-2x$ as $x\rightarrow{0^{-}}$.

Are these thoughts correct, else rectify it please, I can manage right-hand limit after this, just help me with this?

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