Given,

Function $f\left(x\right)=\left\{\begin{array}{cc}\frac{x}{\left|x\right|},& x\ne 0\\ 1,& x=0\end{array}\right., g\left(x\right)=\left\{\begin{array}{cc}\frac{\sin \left(x+1\right)}{\left(x+1\right)},& x\ne -1\\ 1,& x=-1\end{array}\right.$ and $h\left(x\right)=2\left[x\right]-f\left(x\right)$

To find $\underset{x\to 1}{\lim }g\left(h\left(x-1\right)\right)$

Taking $\mathrm{L}.\mathrm{H}.\mathrm{L}$, we take $x=1-h$

$\underset{h\to 0^{-}}{\lim }g\left(h\right(1-h-1\left)\right)$

$\Rightarrow \underset{h\to 0^{-}}{\lim }g\left(2\left[-h\right]-\frac{-h}{\left|-h\right|}\right)$

$\Rightarrow \underset{h\to 0^{-}}{\lim }g\left(2\right(-1)+1)$

$\Rightarrow \underset{h\to 0^{-}}{\lim }g\left(-1\right)=1$

Now taking $\mathrm{R}.\mathrm{H}.\mathrm{L}$, we take $x=1+h$

$\underset{h\to 0^{+}}{\lim }g\left(h\right(1+h-1\left)\right)$

$=\underset{h\to 0^{+}}{\lim }g\left(2\left[h\right]-f\left(h\right)\right)$

$=\underset{h\to 0^{+}}{\lim }g(0-1)$

$\Rightarrow \underset{h\to 0^{+}}{\lim }1=1$