If `f(x) = sin x, x ne npi, n=0, +-1, +-2, +-3=2`, otherwise and `g(x) = x^(2) + 1, x ne 0,2, g(0) =4, g(2) = 5`, then `lim_(x to 0) g[f(x)]` is-

Answer

To solve the problem, we need to find the limit:

\[ \lim_{x \to 0} g[f(x)] \]

where \( f(x) = \sin x \) for \( x \neq n\pi \) (where \( n \) is any integer), and \( f(x) = 2 \) at those points. The function \( g(x) \) is defined as \( g(x) = x^2 + 1 \) for \( x \neq 0, 2 \), with \( g(0) = 4 \) and \( g(2) = 5 \).

### Step 1: Evaluate \( f(x) \) as \( x \to 0 \)

As \( x \) approaches \( 0 \), we find:

\[ f(0) = \sin(0) = 0 \]

Since \( 0 \) is not of the form \( n\pi \) for \( n \neq 0 \), we use the definition of \( f(x) \) as \( \sin x \).

### Step 2: Substitute \( f(x) \) into \( g(x) \)

Now we substitute \( f(x) \) into \( g(x) \):

\[ g[f(x)] = g[\sin x] \]

As \( x \to 0 \), \( \sin x \to 0 \).

### Step 3: Evaluate \( g(0) \)

Now we need to evaluate \( g(0) \):

\[ g(0) = 4 \]

### Step 4: Find the limit

Thus, we can conclude:

\[ \lim_{x \to 0} g[f(x)] = g[0] = 4 \]

### Final Answer

The limit is:

\[ \lim_{x \to 0} g[f(x)] = 4 \]