Let $$f, g$$ and $$h$$ be the real valued functions defined on $$\mathbb{R}$$ as
$$f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}$$, $$g(x) = \begin{cases} \frac{\sin(x+1)}{(x+1)}, & x \neq -1 \\ 1, & x = -1 \end{cases}$$ and $$h(x) = 2[x] - f(x)$$, where $$[x]$$ is the greatest integer $$\leq x$$. Then the value of $$\lim_{x \to 1} g(h(x-1))$$ is

We need to find $$\lim_{x \to 1} g(h(x-1))$$.

Let $$t = x - 1$$, so as $$x \to 1$$, we have $$t \to 0$$.

For $$t \to 0^+$$: $$[t] = 0$$ and $$f(t) = \frac{t}{|t|} = 1$$, so $$h(t) = 0 - 1 = -1$$.

For $$t \to 0^-$$: $$[t] = -1$$ and $$f(t) = \frac{t}{|t|} = -1$$, so $$h(t) = -2 - (-1) = -1$$.

So $$h(t) = -1$$ for all $$t$$ near $$0$$ (but $$t \neq 0$$).

Now, since $$h(t) = -1$$ near $$t = 0$$, we need $$g(-1)$$. From the piecewise definition, $$g(-1) = 1$$.

Hence, $$\lim_{x \to 1} g(h(x-1)) = g(-1) = 1$$.

So, the answer is $$1$$.