Match the following 

	Column I		Column II
A)	$\underset{x\to 2^{+}}{\lim }\left\{x\right\}\frac{\sin (x-2)}{{(x-2)}^2}=$	(p)	– 3
B)	$\underset{x\to 2}{\lim }[2-x]+[x-2]-x=$	(q)	1
C)	$\underset{x\to 2}{\lim }\{2-x\}+\{x-2\}=$	(r)	-1
D)	$\underset{x\to 0^{+}}{\lim }\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}=$	(s)	2

(Note: [ ] is G.I.F and { } is fractional part of function) 

1. A) The value of  $\left\{x\right\}=x-2$ for  x close to 2 and $x>2$

$L=\underset{x\to 2^{+}}{\lim }\frac{(x-2)\sin (x-2)}{{(x-2)}^2}=\underset{x\to 2^{+}}{\lim }\frac{\sin (x-2)}{x-2}=1$

B)  $=\underset{x\to 2}{\lim }[2-x]+[x-2]-x$$=\underset{x\to 2}{\lim }2+[-x]+\left[x\right]-2-x=\underset{x\to 2}{\lim }[-x]+\left[x\right]-x$

$$\begin{gathered} R.H.L=\underset{x\to 2^{+}}{\lim }[-x]+\left[x\right]-2=-3+2-2=-3 \\ L.H.L=\underset{x\to 2^{-}}{\lim }[-x]+\left[x\right]-2=-2+1-2=-3 \\ \therefore L=-3 \end{gathered}$$

C) $\{2-x\}+\{x-2\}=2-x-[2-x]+x-2-[x-2]$

$$\begin{gathered} =-[2-x]-[x-2]=-2-[-x]-\left[x\right]+2=-[-x]-\left[x\right] \\ R.H.L=\underset{x\to 2^{+}}{\lim }-[-x]-\left[x\right]=3-2=1 \\ L.H.L=\underset{x\to 2^{-}}{\lim }-[-x]-\left[x\right]=2-1=1 \\ \therefore L=1 \end{gathered}$$

D)  $L=\underset{x\to 0^{+}}{\lim }\frac{e^{\frac{1}{x}}-e^{\frac{-1}{{ }^x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}=\underset{x\to 0}{\lim }\frac{1-e^{\frac{-2}{x}}}{1+e^{\frac{-2}{x}}}=\frac{1-0}{1+0}=1$