If `l=lim_(xto1^(+))2^(-2^(1/(1-x)))` and `m=lim_(xto1^(+))(x sin (x-[x]))/(x-1)` (where [.] denotes greatest integer function). Then `(l+m)` is ………….

Answer

To solve the problem, we need to evaluate the limits \( l \) and \( m \) separately and then find their sum \( l + m \).

### Step 1: Evaluate \( l = \lim_{x \to 1^+} 2^{-2^{\frac{1}{1-x}}} \)

1. **Substitution**: As \( x \to 1^+ \), let \( x = 1 + h \) where \( h \to 0^+ \). Then, \( 1 - x = -h \).

\[ l = \lim_{h \to 0^+} 2^{-2^{\frac{1}{-h}}} \]

2. **Simplifying the exponent**: The expression becomes:

\[ l = \lim_{h \to 0^+} 2^{-2^{-\frac{1}{h}}} \]

3. **Behavior of the exponent**: As \( h \to 0^+ \), \( -\frac{1}{h} \to -\infty \), thus \( 2^{-\frac{1}{h}} \to 0 \).

4. **Final evaluation**: Therefore, we have:

\[ l = 2^{-0} = 2^0 = 1 \]

### Step 2: Evaluate \( m = \lim_{x \to 1^+} \frac{x \sin(x - [x])}{x - 1} \)

1. **Understanding the greatest integer function**: For \( x \to 1^+ \), \( [x] = 1 \). Thus, \( x - [x] = x - 1 \).

\[ m = \lim_{x \to 1^+} \frac{x \sin(x - 1)}{x - 1} \]

2. **Substitution**: Again, let \( x = 1 + h \) where \( h \to 0^+ \):

\[ m = \lim_{h \to 0^+} \frac{(1 + h) \sin(h)}{h} \]

3. **Breaking down the limit**: This can be rewritten as:

\[ m = \lim_{h \to 0^+} (1 + h) \cdot \lim_{h \to 0^+} \frac{\sin(h)}{h} \]

4. **Using the known limit**: We know that \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):

\[ m = (1 + 0) \cdot 1 = 1 \]

### Step 3: Find \( l + m \)

Now that we have evaluated both limits:

\[ l = 1 \quad \text{and} \quad m = 1 \]

Thus,

\[ l + m = 1 + 1 = 2 \]

### Final Answer

\[ \boxed{2} \]