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EEJ MAIN Mathematics QUESTION #6952
Question 1

For each $x\in\mathbb{R}$, let $[x]$ be the greatest integer $\leq x$. Find: $\displaystyle\lim_{x\to0^-}\dfrac{x([x]+|x|)\sin[x]}{|x|}$

  • $-\sin 1$✔️
  • 1
  • $\sin 1$
Correct Answer Explanation

For $x\to0^-$: $x<0$, $|x|=-x$, $[x]=-1$.

$\dfrac{x((-1)+(-x))\sin(-1)}{-x} = \dfrac{x(-1-x)(-\sin1)}{-x}$

$= \dfrac{x(1+x)\sin1}{x} = (1+x)\sin1$

As $x\to0^-$: limit $= 1\cdot\sin1$... wait: $= \dfrac{x(-1-x)(-\sin1)}{-x} = \dfrac{(1+x)\sin1 \cdot x}{x}=(1+x)\sin1\to\sin1$? Re-doing: numerator $= x(-1-x)\sin(-1)=x(-1-x)(-\sin1)=x(1+x)\sin1$; denominator $=-x$. So $=\dfrac{x(1+x)\sin1}{-x}=-(1+x)\sin1\to-\sin1$. Answer: $\mathbf{-\sin 1}$.