Question

I : If a > 0, then $\underset{x\to \infty }{lim}\frac{[ax+b]}{x}=a$
II : $\underset{x\to \frac{\pi }{2}}{lim}\left[\sin x\right]=0$

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

The correct option is C Both I and II are true

Statement I : If a > 0 then $\underset{x\to \infty }{lim}\frac{[ax+b]}{x}=a$
We have $ax+b-1\le [ax+b]<ax+b$
$a+\frac{b}{x}-\frac{1}{x}\le \frac{[ax+b]}{x}<a+\frac{b}{x}$
$\underset{x\to \infty }{lim}a+\frac{b}{x}-\frac{1}{x}\le \underset{x\to \infty }{lim}\frac{[ax+b]}{x}<\underset{x\to \infty }{lim}a+\frac{b}{x}$
So,$\underset{x\to \infty }{lim}\frac{[ax+b]}{x}=a$
II:$\underset{x\to \frac{\pi }{2}}{lim}\left[\sin x\right]=0$
Here, when $x\to \frac{\pi }{2}$ then $\sin x\to 1$
$\sin x$ is near to 1 but not 1.
So , $\underset{x\to \frac{\pi }{2}}{lim}\left[\sin x\right]=0$