$f\left(x\right)$ $=\left[x\right]+\sqrt{\left\{x\right\}}$ , where [.] and {.} denote the greatest integer function and fractional part respectively, then $f\left(x\right)$  is

$f\left(x\right)$ $=\left[x\right]+\sqrt{\left\{x\right\}}$

$f\left(1\right)=1+0=1$

$f(1+0)=$ $\underset{h\to 0}{\lim }$ $[1+h]+\sqrt{\{1+h\}}$ $=\underset{h\to 0}{\lim }1+$ $\sqrt{h}=1$

$f(1-0)$ $=\underset{h\to 0}{\lim }$ $[1-h]+\sqrt{\{1-h\}}$ $=\underset{h\to 0}{\lim }$ $0+\sqrt{1-h}=1$

$\therefore$  $f\left(x\right)$  is continuous at $x=1$

$Lf\text{'}\left(1\right)$ $=\underset{h\to 0}{\lim }$ $\frac{[1-h]+\sqrt{\{1-h\}}-f\left(1\right)}{–h}$

             $=–\underset{h\to 0}{\lim }\frac{0+\sqrt{1-h}-1}{h}$

             $=–\underset{h\to 0}{\lim }\frac{(1+h)-1}{h(\sqrt{1-h}+1)}=1/2$

$Rf\text{'}\left(1\right)$ $=\underset{h\to 0}{\lim }$ $\frac{[1+h]+\sqrt{\{1+h\}}-f\left(1\right)}{h}$

             $=\underset{h\to 0}{\lim }$ $\frac{1+\sqrt{h}–1}{h}=+\infty$

$\therefore$  At $x=1$ , $f\left(x\right)$  is continuous but not differentiable.