The value of $$\lim_{x \to 0} \left[ (1 - e^x) \frac{\sin x}{|x|} \right]$$ equals:
Step-by-Step Solution
Key Concept: Study the quantity inside the greatest integer function from both sides of \(0\):
\[
(1-e^x)\frac{\sin x}{|x|}.
\]
Although this expression tends to \(0\), it tends to \(0\) through negative values from both sides. Therefore its greatest integer value is \(-1\) near \(0\), not \(0\).
\subsection*{Question 4: Solution}
Consider
\[
A(x)=(1-e^x)\frac{\sin x}{|x|}.
\]
As \(x\to 0\),
\[
1-e^x\sim -x.
\]
For \(x>0\),
\[
\frac{\sin x}{|x|}
=\frac{\sin x}{x}\to 1.
\]
Thus
\[
A(x)\sim -x<0.
\]
So \(A(x)\) approaches \(0\) through negative values from the right.
For \(x<0\),
\[
|x|=-x,
\]
and
\[
\frac{\sin x}{|x|}
\sim \frac{x}{-x}=-1.
\]
Also,
\[
1-e^x\sim -x>0.
\]
Hence
\[
A(x)\sim (-x)(-1)=x<0.
\]
So \(A(x)\) approaches \(0\) through negative values from the left as well.
Therefore, for all sufficiently small non-zero \(x\),
\[
-1<A(x)<0.
\]
Hence
\[
[A(x)]=-1.
\]
Thus
\[
\lim_{x\to 0}
\left[
(1-e^x)\frac{\sin x}{|x|}
\right]
=-1.
\]
Correct Answer: 2