Question

Let $[x]$ denote the greatest integer less than or equal to x. Then $\lim_{x\rightarrow0^{-}}\frac{\sin[x]}{[x]}$ is equal to

• A. 1
• B. $-\sin1$
• C. -2
• D. 0
• E. $\sin1$

Hint:

Logic Tip: Do not confuse this with the standard limit identity $\lim_{x\to0}\frac{\sin x}{x} = 1$. The presence of the greatest integer function $[x]$ transforms the continuous variable into discrete constants depending on the direction of approach!

Answer 1

Answer: (E)

Concept:
Calculus - Left-Hand Limits and the Greatest Integer Function.
Step 1: Analyze the given limit condition.
We need to find the Left-Hand Limit (LHL) as $x$ approaches $0$. $$ \lim_{x\rightarrow0^{-}} $$ This indicates that $x$ takes values that are very close to $0$, but strictly less than $0$ (e.g., $x = -0.001$).
Step 2: Evaluate the Greatest Integer Function $[x]$.
For any value of $x$ in the interval $-1 \le x<0$, the greatest integer less than or equal to $x$ is $-1$. Since $x \rightarrow 0^{-}$, it falls perfectly into this interval. $$ \text{As } x \rightarrow 0^{-}, \quad [x] = -1 $$
Step 3: Substitute $[x]$ into the limit expression.
Replace all instances of $[x]$ in the expression with the constant $-1$: $$ \lim_{x\rightarrow0^{-}}\frac{\sin[x]}{[x]} = \frac{\sin(-1)}{-1} $$
Step 4: Simplify the trigonometric expression.
Recall that sine is an odd function, meaning $\sin(-\theta) = -\sin(\theta)$. $$ \frac{-\sin(1)}{-1} $$ The negative signs in the numerator and denominator cancel out: $$ \sin(1) $$