JEE Main 2023 — Limits Question with Solution

We have,

$\underset{x\to a}{\lim }\left(\left[x-5\right]-\left[2x+2\right]\right)=0$

$\Rightarrow \underset{x\to a}{\lim }\left(\left[x\right]-5-\left[2x\right]-2\right)=0$

$\Rightarrow \underset{x\to a}{\lim }\left(\left[x\right]-\left[2x\right]-7\right)=0$

$\Rightarrow \underset{x\to a}{\lim }\left(\left[x\right]-\left[2x\right]\right)=7$

Now when $a\in I$ (integer) then $a-2a=7\Rightarrow a=-7$

Now when $a\in I+f$ where $f$ is fraction,

So, $\left[I+f\right]-\left[2I+2f\right]=7$

$\Rightarrow I+\left[f\right]-2I-\left[2f\right]=7$

$\Rightarrow I+0-2I-\left[2f\right]=7$

$\Rightarrow -I-\left[2f\right]=7$

Now taking case $\left(1\right)$ when $f\in \left(0,\frac{1}{2}\right)$

Then $\Rightarrow -I-0=7\Rightarrow I=-7\Rightarrow a\in \left(-7,-6.5\right)$

Now taking case $\left(2\right)$ when $f\in \left(\frac{1}{2},1\right)$

So, $-I-1=7\Rightarrow I=-8\Rightarrow a\in \left(-7.5,7\right)$

Hence, from both the cases we can say that $a\in \left(-7.5,-6.5\right)$.