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Is the following limit exist or not? $$\lim_{x\to 0} {\frac{\lfloor \frac{3}{2} +x\rfloor }{x}}$$

I have no idea about find right-hand and left-hand limits.

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    $\begingroup$ What is $[[ ]]$ $\endgroup$ Jan 25, 2017 at 20:48
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    $\begingroup$ My first thought is the floor function, but that's clearly wrong... $\endgroup$
    – y_prime
    Jan 25, 2017 at 20:50
  • $\begingroup$ Yes, floor function. $\endgroup$
    – Domates
    Jan 25, 2017 at 20:51

1 Answer 1

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The limit is not defined (it's infinity). Just fill in $x = 0$ and you get $\lim_{x\to0} \frac{\lfloor\frac{3}{2}+x\rfloor}{x} = \frac{\lfloor\frac{3}{2}+0\rfloor}{0} = \frac{1}{0}=\infty$. Note: for $x\to0$ with $x<0$, the limit is $-\infty$, for $x>0$ is it $+\infty$ You can also see this clearly when plotting the graph: enter image description here

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    $\begingroup$ Isn't $\lfloor 3/2+0\rfloor =1$? $\endgroup$
    – kingW3
    Jan 25, 2017 at 21:19
  • $\begingroup$ Yes indeed, I was a bit too fast there :-) $\endgroup$
    – mitchbus
    Jan 25, 2017 at 21:20

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