The value of `underset(|x| rarr oo)("lim") cos (tan^(-1) (sin (tan^(-1) x)))` is equal to

Options

A. `-1`
B. `sqrt2`
C. `- 1/sqrt2`
D. `1/sqrt2`

Answer

To find the value of the limit

\[ \lim_{|x| \to \infty} \cos\left(\tan^{-1}\left(\sin\left(\tan^{-1} x\right)\right)\right), \]

we will proceed step by step.

### Step 1: Substitute \( \tan^{-1} x \)
Let \( \theta = \tan^{-1} x \). Then, we have:

\[ x = \tan \theta. \]

### Step 2: Find \( \sin(\tan^{-1} x) \)
Using the identity for sine in terms of tangent, we can express \( \sin(\theta) \):

\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}}. \]

Thus, we have:

\[ \sin(\tan^{-1} x) = \frac{x}{\sqrt{x^2 + 1}}. \]

### Step 3: Substitute back into the limit
Now, we need to find \( \tan^{-1}(\sin(\tan^{-1} x)) \):

\[ \tan^{-1}\left(\frac{x}{\sqrt{x^2 + 1}}\right). \]

### Step 4: Find \( \tan(\tan^{-1}(\sin(\tan^{-1} x))) \)
Let \( \phi = \tan^{-1}\left(\frac{x}{\sqrt{x^2 + 1}}\right) \). Then,

\[ \tan(\phi) = \frac{x}{\sqrt{x^2 + 1}}. \]

### Step 5: Find \( \cos(\tan^{-1}(\sin(\tan^{-1} x))) \)
We can express \( \cos(\phi) \) using the identity:

\[ \cos(\phi) = \frac{1}{\sqrt{1 + \tan^2(\phi)}}. \]

Substituting \( \tan(\phi) \):

\[ \cos(\phi) = \frac{1}{\sqrt{1 + \left(\frac{x}{\sqrt{x^2 + 1}}\right)^2}} = \frac{1}{\sqrt{1 + \frac{x^2}{x^2 + 1}}}. \]

### Step 6: Simplify the expression
Now, simplify the expression inside the square root:

\[ 1 + \frac{x^2}{x^2 + 1} = \frac{x^2 + 1 + x^2}{x^2 + 1} = \frac{2x^2 + 1}{x^2 + 1}. \]

Thus, we have:

\[ \cos(\phi) = \frac{1}{\sqrt{\frac{2x^2 + 1}{x^2 + 1}}} = \frac{\sqrt{x^2 + 1}}{\sqrt{2x^2 + 1}}. \]

### Step 7: Evaluate the limit
Now we need to evaluate the limit:

\[ \lim_{|x| \to \infty} \frac{\sqrt{x^2 + 1}}{\sqrt{2x^2 + 1}}. \]

Dividing numerator and denominator by \( |x| \):

\[ = \lim_{|x| \to \infty} \frac{\sqrt{1 + \frac{1}{x^2}}}{\sqrt{2 + \frac{1}{x^2}}}. \]

As \( |x| \to \infty \), \( \frac{1}{x^2} \to 0 \):

\[ = \frac{\sqrt{1 + 0}}{\sqrt{2 + 0}} = \frac{1}{\sqrt{2}}. \]

### Final Answer
Thus, the value of the limit is:

\[ \frac{1}{\sqrt{2}}. \]