Let `f(x)=lim_(ntooo)(cos sqrt(x/n))^(n),g(x)=lim_(nto oo)(1-x+x root(n)(e))^(n)`
Now consider the function `y=h(x)` where `h(x)=tan^(-1)(g^(-1)f^(-1)(x))`.
Range of the function `y=h(x)` is

Options

A. `(0,(pi)/2)`
B. `(-(pi)/2,0)`
C. `R`
D. `(-(pi)/2,(pi)/2)`

Answer

To solve the problem step by step, we will find the functions \( f(x) \) and \( g(x) \) first, and then we will analyze the function \( h(x) \) to determine its range.

### Step 1: Finding \( f(x) \)

Given:
\[ f(x) = \lim_{n \to \infty} \left( \cos\left(\sqrt{\frac{x}{n}}\right) \right)^n \]

As \( n \to \infty \), \( \sqrt{\frac{x}{n}} \to 0 \). Therefore, we can use the limit:
\[ \cos\left(\sqrt{\frac{x}{n}}\right) \to \cos(0) = 1 \]

This is an indeterminate form \( 1^\infty \). We can rewrite it using the exponential function:
\[ f(x) = \lim_{n \to \infty} e^{n \ln\left(\cos\left(\sqrt{\frac{x}{n}}\right)\right)} \]

Using the approximation \( \cos(y) \approx 1 - \frac{y^2}{2} \) for small \( y \):
\[ \cos\left(\sqrt{\frac{x}{n}}\right) \approx 1 - \frac{\left(\sqrt{\frac{x}{n}}\right)^2}{2} = 1 - \frac{x}{2n} \]

Thus,
\[ \ln\left(\cos\left(\sqrt{\frac{x}{n}}\right)\right) \approx -\frac{x}{2n} \]

Substituting this back:
\[ f(x) = \lim_{n \to \infty} e^{n \left(-\frac{x}{2n}\right)} = e^{-\frac{x}{2}} \]

### Step 2: Finding \( f^{-1}(x) \)

To find \( f^{-1}(x) \):
\[ y = e^{-\frac{x}{2}} \implies -\frac{x}{2} = \ln(y) \implies x = -2 \ln(y) \]
Thus,
\[ f^{-1}(x) = -2 \ln(x) \]

### Step 3: Finding \( g(x) \)

Given:
\[ g(x) = \lim_{n \to \infty} \left(1 - x + x e^{\frac{1}{n}}\right)^n \]

As \( n \to \infty \), \( e^{\frac{1}{n}} \to 1 \):
\[ g(x) = \lim_{n \to \infty} \left(1 - x + x\right)^n = 1^n = 1 \]

This is also an indeterminate form \( 1^\infty \). We can rewrite it:
\[ g(x) = \lim_{n \to \infty} e^{n \ln\left(1 - x + x e^{\frac{1}{n}}\right)} \]

Using the approximation \( \ln(1 + y) \approx y \) for small \( y \):
\[ \ln\left(1 - x + x e^{\frac{1}{n}}\right) \approx \ln(1 - x + x) = \ln(1) = 0 \]

Thus,
\[ g(x) = e^{\lim_{n \to \infty} n \cdot 0} = e^0 = 1 \]

### Step 4: Finding \( g^{-1}(x) \)

Since \( g(x) = e^x \), we find:
\[ g^{-1}(x) = \ln(x) \]

### Step 5: Finding \( h(x) \)

Now we can find \( h(x) \):
\[ h(x) = \tan^{-1}(g^{-1}(f^{-1}(x))) = \tan^{-1}(\ln(-2 \ln(x))) \]

### Step 6: Finding the Range of \( h(x) \)

To find the range of \( h(x) \):
1. The argument \( -2 \ln(x) \) is defined for \( x > 0 \).
2. As \( x \to 0^+ \), \( -2 \ln(x) \to \infty \) and \( \ln(-2 \ln(x)) \to \infty \).
3. As \( x \to 1 \), \( -2 \ln(1) = 0 \) and \( \ln(0) \to -\infty \).

Thus, as \( x \) varies from \( 0 \) to \( 1 \), \( h(x) \) will vary from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

### Final Range of \( h(x) \)

The range of the function \( y = h(x) \) is:
\[ \boxed{\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)} \]