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Problem. Let $a$ be an integer such that $$ \lim_{x \to 7}\frac{18-[1-x]}{[x-3a]} $$ exists where $[t]$ is the greatest integer function. What should be that value of $a$?

My approach was divided in three steps:

  1. Directly put $x=7$. That gave me the expression

    $$\frac{24}{7-3a} \tag{1} $$

  2. Approaching from positive direction that is taking $x=h+7$ where $h$ is tending to zero.

    $$\frac{25}{7-3a} \tag{2} $$

  3. Approaching from negative direction, i.e. $x=7-h$ where $h$ is tending to zero.

    $$\frac{24}{7-3a-1} \tag{3} $$

Now if I equate the expression $\text{(2)}$ and $\text{(3)}$ I will get $a=-6$, but that would give a different value at expression $\text{(1)}$.

Is the limit non existent here, because for a limit to exist $\text{R.H.L} = \text{L.H.L} = f(a)$ [$\text{R.H.L} = $ Right hand limit] or we just ignore the $f(a)$ condition.

Sangchul Lee
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VMnM7
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    In computing $\lim_{x\to c}F(x)$ we are not interested in the value $F(c)$, nor do we care if that even exists. Instead, we are interested in the values of $F(x)$ where $x$ is near $c$, but not actually equal to $c$. – lulu Jul 05 '22 at 18:37
  • Is that applicable in every case. Because I thought it applies only when $f(c)$ is undefined like $\infty$ – VMnM7 Jul 05 '22 at 19:00
  • It is applicable in every case. The limit exists if and only if the two one-sided limits both exist and are equal, the value at the limit point is irrelevant. – lulu Jul 05 '22 at 19:03

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