Question

Match the following:

Here $[.]$ and $\left\{x\right\}$ are respectively greatest integer function and fractional part functions

Answer 1

(A) $\frac{\tan x}{x}>1$ and $\frac{\tan x}{x}\to 1$ for $x\to 0$$\Rightarrow \frac{\tan x}{x}=1+h;h\to 0^{+}$

$\Rightarrow 200\frac{\tan x}{x}=200+200h;h\to 0^{+}$$\Rightarrow \left[200\frac{\tan x}{x}\right]=200$

Also $\frac{\sin x}{x}<1$ and $\frac{\sin x}{x}\to 1$ for $x\to 0$$\Rightarrow \frac{\sin x}{x}=1-h;h\to 0^{+}$

$\Rightarrow 200\left(\frac{\sin x}{x}\right)=200-200h;h\to 0^{+}$$\Rightarrow \left[200\frac{\sin x}{x}\right]=199$

$\therefore \underset{x\to 0}{lim}(\left[200\frac{\tan x}{x}\right]+\left[200\frac{\sin x}{x}\right])=399$

(B) $100\frac{\tan x}{x}=100(1+h)=100+100h;h\to 0^{+}$

$\therefore \left\{100\frac{\tan x}{x}\right\}=100h\to 0$ as $h\to 0^{+}$

Also $\left\{100\frac{\sin x}{x}\right\}=\left\{100(1-h)\right\}=\{100-100h\}$

$=\{99+(1-100h)\}=1-100h\to 1h\to 0^{+}$

$\therefore \underset{x\to 0}{lim}\left\{100\frac{\tan x}{x}\right\}+\left\{100\frac{\sin x}{x}\right\}=0+1=1$

(C) $\frac{\sin x}{x}=1-h;h\to 0^{+}$ as $x\to 0^{+}$

$\Rightarrow \left[\frac{\sin x}{x}\right]=[1-h]=0$ and $\frac{x}{{\tan }^{-1}x}>1$ for $4x\to 0$

$\Rightarrow \frac{x}{{\tan }^{-1}x}=1+h;h\to 0^{+}$

$\Rightarrow \left[\frac{x}{{\tan }^{-1}x}\right]=1$

$\therefore \underset{x\to 0}{lim}(100\left[\frac{\sin x}{x}\right]+200\left[\frac{x}{{\tan }^{-1}x}\right])=0+200=200$

(D) As $x\to 0,\frac{{\sin }^{-1}x}{x}>1$ and $\frac{{\sin }^{-1}x}{x}\to 1$ as $x\to 0$

$\Rightarrow \frac{{\sin }^{-1}x}{x}=1+h;h\to 0^{+}$

$\therefore \left\{\frac{{\sin }^{-1}x}{x}\right\}=h\to 0$ as $h\to 0^{+}$

Also $\frac{{\tan }^{-1}x}{x}<1$ and $\frac{{\tan }^{-1}x}{x}\to 1$ as $x\to 0$

$\Rightarrow 200\frac{{\tan }^{-1}x}{x}=200(1-h);h\to 0^{+}$

$\Rightarrow \left[200\frac{{\tan }^{-1}x}{x}\right]=[200-200h]=199$

$\therefore \underset{x\to 0}{lim}\left\{\frac{{\sin }^{-1}x}{x}\right\}+\left[200\frac{{\tan }^{-1}x}{x}\right]=0+199=199$