`lim_(xto oo) (int_(0)^(x)tan^(-1)dt)/(sqrt(x^(2)+1))` is equal to

Options

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

Explanation

To solve the limit

\[ \lim_{x \to \infty} \frac{\int_{0}^{x} \tan^{-1}(t) \, dt}{\sqrt{x^2 + 1}}, \]

we will follow these steps:

### Step 1: Evaluate the integral
We need to evaluate the integral \(\int_{0}^{x} \tan^{-1}(t) \, dt\). We can use integration by parts for this. Let:
- \(u = \tan^{-1}(t)\) and \(dv = dt\).
Then, we have:
- \(du = \frac{1}{1+t^2} dt\) and \(v = t\).

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du, \]
we get:
\[ \int_{0}^{x} \tan^{-1}(t) \, dt = \left[ t \tan^{-1}(t) \right]_{0}^{x} - \int_{0}^{x} \frac{t}{1+t^2} dt. \]

### Step 2: Evaluate the boundary terms
Now, we evaluate the boundary terms:
\[ \left[ t \tan^{-1}(t) \right]_{0}^{x} = x \tan^{-1}(x) - 0 = x \tan^{-1}(x). \]

### Step 3: Evaluate the second integral
Next, we need to evaluate \(\int_{0}^{x} \frac{t}{1+t^2} dt\). We can use the substitution \(u = 1 + t^2\), which gives \(du = 2t \, dt\) or \(dt = \frac{du}{2t}\). The limits change from \(t = 0\) to \(t = x\) which translates to \(u = 1\) to \(u = 1 + x^2\). Thus,
\[ \int_{0}^{x} \frac{t}{1+t^2} dt = \frac{1}{2} \int_{1}^{1+x^2} \frac{1}{u} du = \frac{1}{2} \left[ \ln(u) \right]_{1}^{1+x^2} = \frac{1}{2} \left( \ln(1+x^2) - \ln(1) \right) = \frac{1}{2} \ln(1+x^2). \]

### Step 4: Combine results
Combining these results, we have:
\[ \int_{0}^{x} \tan^{-1}(t) \, dt = x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2). \]

### Step 5: Substitute back into the limit
Now substituting back into the limit:
\[ \lim_{x \to \infty} \frac{x \tan^{-1}(x) - \frac{1}{2} \ln(1+x^2)}{\sqrt{x^2 + 1}}. \]

### Step 6: Simplify the limit
As \(x \to \infty\), \(\tan^{-1}(x) \to \frac{\pi}{2}\), so:
\[ x \tan^{-1}(x) \sim \frac{\pi}{2} x. \]
Also, \(\sqrt{x^2 + 1} \sim x\).

Thus, the limit simplifies to:
\[ \lim_{x \to \infty} \frac{\frac{\pi}{2} x - \frac{1}{2} \ln(1+x^2)}{x}. \]

### Step 7: Separate the terms
This can be separated as:
\[ \lim_{x \to \infty} \left( \frac{\pi}{2} - \frac{\frac{1}{2} \ln(1+x^2)}{x} \right). \]

As \(x \to \infty\), \(\frac{\ln(1+x^2)}{x} \to 0\) because \(\ln(1+x^2) \sim 2 \ln(x)\) and \(\frac{2 \ln(x)}{x} \to 0\).

### Step 8: Final result
Thus, the limit evaluates to:
\[ \frac{\pi}{2}. \]

### Conclusion
Therefore, the final result is:
\[ \lim_{x \to \infty} \frac{\int_{0}^{x} \tan^{-1}(t) \, dt}{\sqrt{x^2 + 1}} = \frac{\pi}{2}. \]