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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.

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

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 $2x+1$.

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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You are correct: for $x\in (-1/2,0)$, $\{2x\}=2x-\lfloor 2x\rfloor=2x+1$, and as $x\to 0^-$, $$\dfrac{e^{\lfloor x\rfloor}+|x|-1}{\lfloor x\rfloor+\{2x\}}= \dfrac{e^{-1}-x-1}{-1+2x+1}\to \dfrac{\overbrace{e^{-1}-1}^{<0}}{0^-}=+\infty.$$

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  • $\begingroup$ why not $-\infty$? $\endgroup$ – mnulb May 16 '18 at 14:32
  • $\begingroup$ Because $e^{-1}-1<0$. $\endgroup$ – Robert Z May 16 '18 at 14:33
  • $\begingroup$ @mnulb Any further doubt? $\endgroup$ – Robert Z May 16 '18 at 14:40
  • $\begingroup$ no, you cleared all, thanks! $\endgroup$ – mnulb May 16 '18 at 14:41

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