Let $l=\underset{x\to 2}{\lim }\frac{\left[x\right]}{\left\{x\right\}}$  and $m=\underset{x\to 2}{\lim }\frac{\left\{x\right\}}{\left[x\right]}$  where $\left[y\right]$  and  $\left\{y\right\}$ denotes largest integer less than or equal to $y$  and fractional part of $y$  respectively.  Then which one of the following alternative is correct?

$\underset{x\to 2^{+}}{\lim }\frac{\left[x\right]}{\left\{x\right\}}=\frac{\left[x\right]}{x-2}=\underset{h\to 0}{\lim }\frac{[2+h]}{h}=\underset{h\to 0}{\lim }\frac{2}{h}=+\infty \because \left\{x\right\}=x-\left[x\right]$        
$\underset{x\to 2^{-}}{\lim }\frac{\left[x\right]}{\left\{x\right\}}=\frac{\left[x\right]}{x-1}=\underset{h\to 0}{\lim }\frac{[2-h]}{(2-h)-1}=\frac{1}{1}=1$
$\therefore l$ does not exist.
Now  $\underset{x\to 2^{+}}{\lim }\frac{\left\{x\right\}}{\left[x\right]}=\underset{x\to 2^{+}}{\lim }\frac{x-2}{\left[x\right]}=\underset{h\to 0}{\lim }\frac{h}{[2+h]}=\underset{x\to 0}{\lim }\frac{h}{2}=0$
and  $\underset{x\to 2^{-}}{\lim }\frac{\left\{x\right\}}{\left[x\right]}=\underset{x\to 2^{-}}{\lim }\frac{x-1}{\left[x\right]}=\underset{h\to 0}{\lim }\frac{1-h}{[2-h]}=1$
$\therefore m$ does not exist.