.Let `f,g` and `h` be the real valued function defined on `R` as `f(x)={(x/absx,xne0),(1,x=0):} , g(x)={(sin(x+1)/(x+1),xne0),(1,x=-1):}` and `h(x)=2[x]-f(x),` where `[.]` is the greatest integer `lex`.Then the value of `lim_(x rarr1)g(h(x-1))` is

Options

A. `1`
B. `0`
C. `sin (1)`
D. `-1`

Answer

To solve the problem, we need to analyze the functions \( f(x) \), \( g(x) \), and \( h(x) \) step by step and then evaluate the limit \( \lim_{x \to 1} g(h(x-1)) \).

### Step 1: Define the functions
The functions are defined as follows:
- \( f(x) = \begin{cases} \frac{x}{|x|} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \)

- \( g(x) = \begin{cases} \frac{\sin(x+1)}{x+1} & \text{if } x \neq -1 \\ 1 & \text{if } x = -1 \end{cases} \)

- \( h(x) = 2 \lfloor x \rfloor - f(x) \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \).

### Step 2: Evaluate \( h(x-1) \) as \( x \to 1 \)
When \( x \to 1 \), we have \( x - 1 \to 0 \). Therefore, we need to evaluate \( h(0) \):
- \( \lfloor 0 \rfloor = 0 \)
- \( f(0) = 1 \)

Thus,
\[ h(0) = 2 \lfloor 0 \rfloor - f(0) = 2 \cdot 0 - 1 = -1 \]

### Step 3: Evaluate \( g(h(x-1)) \)
Now we need to evaluate \( g(h(0)) = g(-1) \):
- From the definition of \( g(x) \), when \( x = -1 \), we have \( g(-1) = 1 \).

### Step 4: Evaluate the limit
Now we can find the limit:
\[ \lim_{x \to 1} g(h(x-1)) = g(h(0)) = g(-1) = 1 \]

### Conclusion
The value of \( \lim_{x \to 1} g(h(x-1)) \) is \( 1 \).