Question

STATEMENT-1: $\left[\underset{x\to 0}{lim}\frac{\sin x}{x}\right]=0$
STATEMENT-2: For $x\in (-\delta ,\delta )$, where $\delta$ is positive and $\delta \to 0,\tan x>x.$

• A. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
• B. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is not the correct explanation for STATEMENT-1
• C. STATEMENT-1 is True, STATEMENT-2 is False
• D. STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

The correct option is B STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is not the correct explanation for STATEMENT-1

For $x\in (-\delta ,\delta ),sinx<x$ or ${lim}_{x\to 0}\frac{\sin x}{x}=1^{-}$ or $\left[\underset{x\to 0}{lim}\frac{\sin x}{x}\right]=0$
Also, $x\in (-\delta ,\delta ),\tan x>x.$ But from this, nothing can be said about the relation between $\sin x$ and $x$.
Hence, both the statements are true but statement 2 is not the correct explanation of statement 1.