Let l = mathop {lim}limits_{x to 0} frac{[x]^ | vedclass

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(B) For $l = \mathop {\lim}\limits_{x 	o 0} \frac{[x]^2}{x^2}$:As $x 	o 0^+$,$[x] = 0$,so $\frac{[x]^2}{x^2} = 0$. Thus,$\mathop {\lim}\limits_{x

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$m$ exists but $l$ does not

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(B) For $l = \mathop {\lim}\limits_{x 	o 0} \frac{[x]^2}{x^2}$:As $x 	o 0^+$,$[x] = 0$,so $\frac{[x]^2}{x^2} = 0$. Thus,$\mathop {\lim}\limits_{x 	o 0^+} \frac{[x]^2}{x^2} = 0$.As $x 	o 0^-$,$[x] = -1$,so $\frac{[x]^2}{x^2} = \frac{1}{x^2} 	o \infty$. Thus,$l$ does not exist.For $m = \mathop {\lim}\limits_{x 	o 0} \frac{[x^2]}{x^2}$:As $x 	o 0$,$x^2$ is a small positive value approaching $0$ from the right $(0 Therefore,$[x^2] = 0$ for all $x 
eq 0$ in the neighborhood of $0$.Thus,$\mathop {\lim}\limits_{x 	o 0} \frac{0}{x^2} = 0$. So,$m$ exists and $m = 0$.Conclusion: $m$ exists but $l$ does not.

Let $l = \mathop {\lim}\limits_{x \to 0} \frac{[x]^2}{x^2}$ and $m = \mathop {\lim}\limits_{x \to 0} \frac{[x^2]}{x^2}$,where $[ \cdot ]$ denotes the greatest integer function. Then:

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