Let `lim_(x->oo) sqrt((x-cos^2x)/(x+sinx)) and I_2 = lim_(h->0^+) int_(-1)^1 hdx/(h^2+x^2)`. Then

Options

A. Both `l _(1) and l _(2)` are less than 22/7
B. One of the two limits is rational and other irrational
C. `l _(2) gt l _(1)`
D. `l _(2)` is greater than 3 times of `l _(1)`

Answer

To solve the given problem, we need to evaluate two limits:

1. \( I_1 = \lim_{x \to \infty} \sqrt{\frac{x - \cos^2 x}{x + \sin x}} \)
2. \( I_2 = \lim_{h \to 0^+} \int_{-1}^{1} \frac{h \, dx}{h^2 + x^2} \)

### Step 1: Evaluate \( I_1 \)

We start with the limit:

\[ I_1 = \lim_{x \to \infty} \sqrt{\frac{x - \cos^2 x}{x + \sin x}} \]

**Hint:** Factor out \( x \) from both the numerator and the denominator.

\[ I_1 = \lim_{x \to \infty} \sqrt{\frac{x(1 - \frac{\cos^2 x}{x})}{x(1 + \frac{\sin x}{x})}} = \lim_{x \to \infty} \sqrt{\frac{1 - \frac{\cos^2 x}{x}}{1 + \frac{\sin x}{x}}} \]

**Hint:** As \( x \) approaches infinity, \( \frac{\cos^2 x}{x} \) and \( \frac{\sin x}{x} \) both approach 0.

\[ I_1 = \sqrt{\frac{1 - 0}{1 + 0}} = \sqrt{1} = 1 \]

### Step 2: Evaluate \( I_2 \)

Next, we evaluate the second limit:

\[ I_2 = \lim_{h \to 0^+} \int_{-1}^{1} \frac{h \, dx}{h^2 + x^2} \]

**Hint:** Recognize that the integral can be simplified by factoring out \( h \).

\[ I_2 = \lim_{h \to 0^+} h \int_{-1}^{1} \frac{dx}{h^2 + x^2} \]

**Hint:** The integral \( \int \frac{dx}{h^2 + x^2} \) can be solved using the formula \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \).

\[ \int_{-1}^{1} \frac{dx}{h^2 + x^2} = \left[ \frac{1}{h} \tan^{-1}\left(\frac{x}{h}\right) \right]_{-1}^{1} \]

**Hint:** Evaluate the limits of the arctangent function.

\[ = \frac{1}{h} \left( \tan^{-1}\left(\frac{1}{h}\right) - \tan^{-1}\left(\frac{-1}{h}\right) \right) \]

**Hint:** Use the property of the arctangent function, \( \tan^{-1}(-x) = -\tan^{-1}(x) \).

\[ = \frac{1}{h} \left( \tan^{-1}\left(\frac{1}{h}\right) + \tan^{-1}\left(\frac{1}{h}\right) \right) = \frac{2}{h} \tan^{-1}\left(\frac{1}{h}\right) \]

**Hint:** As \( h \to 0^+ \), \( \tan^{-1}\left(\frac{1}{h}\right) \) approaches \( \frac{\pi}{2} \).

\[ I_2 = \lim_{h \to 0^+} h \cdot \frac{2}{h} \cdot \frac{\pi}{2} = \lim_{h \to 0^+} \pi = \pi \]

### Conclusion

Now we have:

- \( I_1 = 1 \)
- \( I_2 = \pi \)

### Final Answer

We can compare \( I_1 \) and \( I_2 \):

- \( I_1 < \frac{22}{7} \) (since \( \pi \approx 3.14 \))
- \( I_2 = \pi \) (which is irrational)
- \( I_2 > 3 \times I_1 \) (since \( \pi > 3 \))