`Lt_(x rarr oo) x tan^(-1)"" (2)/(x)` equal to

Options

A. 1
B. `(1)/(2)`
C. `(1)/(4)`
D. 2

Answer

To solve the limit \( \lim_{x \to \infty} x \tan^{-1}\left(\frac{2}{x}\right) \), we can follow these steps:

### Step 1: Substitute \( x \) with \( \frac{1}{t} \)
We start by substituting \( x = \frac{1}{t} \). As \( x \to \infty \), \( t \to 0 \). Thus, we rewrite the limit:

\[ \lim_{x \to \infty} x \tan^{-1}\left(\frac{2}{x}\right) = \lim_{t \to 0} \frac{1}{t} \tan^{-1}(2t) \]

### Step 2: Rewrite the limit
Now, we can express the limit as:

\[ \lim_{t \to 0} \frac{\tan^{-1}(2t)}{t} \]

### Step 3: Multiply and divide by 2
To use the known limit property, we multiply and divide by 2:

\[ \lim_{t \to 0} \frac{\tan^{-1}(2t)}{t} = \lim_{t \to 0} \frac{\tan^{-1}(2t)}{2t} \cdot 2 \]

### Step 4: Apply the limit property
We know from limit properties that:

\[ \lim_{u \to 0} \frac{\tan^{-1}(u)}{u} = 1 \]

In our case, \( u = 2t \). As \( t \to 0 \), \( 2t \to 0 \) as well. Therefore:

\[ \lim_{t \to 0} \frac{\tan^{-1}(2t)}{2t} = 1 \]

### Step 5: Final calculation
Now substituting back into our limit, we have:

\[ \lim_{t \to 0} \frac{\tan^{-1}(2t)}{t} = 2 \cdot 1 = 2 \]

### Conclusion
Thus, the final answer is:

\[ \lim_{x \to \infty} x \tan^{-1}\left(\frac{2}{x}\right) = 2 \]