Let f, g and h be the real valued functions defined on ℝ as f(x) =...

Given,

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

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

Taking $L . H . L$, we take $x = 1 - h$

$\lim_{h \to {0}^{-}}g ( h ( 1 - h - 1 ) )$

$⇒ \lim_{h \to {0}^{-}}g\left(2\left[- h\right] - \dfrac{- h}{\left|- h\right|}\right)$

$⇒ \lim_{h \to {0}^{-}}g ( 2 ( - 1 ) + 1 )$

$⇒ \lim_{h \to {0}^{-}}g\left(- 1\right) = 1$

Now taking $R . H . L$, we take $x = 1 + h$

$\lim_{h \to {0}^{+}}g ( h ( 1 + h - 1 ) )$

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

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

$⇒ \lim_{h \to {0}^{+}}1 = 1$